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TEXT-BOOK 


OF 


GEOMETEY 


REVISED    EDITION. 


BY 


G.  A.  WENTWORTH,  A.M., 

AUTHOR  OF  A   SERIES  OF  TEXT-BOOKS   IN    MATHEMATICS. 


BOSTON,  U.S.A.: 

GINN   &   COMPANY,   PUBLTSHEKS. 

1893. 


0/  n 


Entered,  according  to  Act  of  Congress,  in  the  yearr  1888,  by 

G.  A.  WENTWORTH, 
In  the  OflSce  of  the  Librarian  of  Congress,  at  Washington. 


All  Rights  Rbsebvbd. 


Typography  by  J.  8.  Cushing  &  Co.,  Boston,  U.S.A. 


Prbsswork  by  Ginn  &  Co..  Boston,  U.S.A. 


PREFACE. 


ni  T'OST  persons  do  not  possess,  and  do  not  easily  acquire,  the  power 
-^^-*-  of  abstraction  requisite  for  apprehending  geometrical  concep- 
tions, and  for  keeping  in  mind  the  successive  steps  of  a  continuous 
argument.  Hence,  with  a  very  large  proportion  of  beginners  in  Geom- 
etry, it  depends  mainly  upon  the  form  in  which  the  subject  is  pre- 
sented whether  they  pursue  the  study  with  indifference,  not  to  say 
aversion,  or  with  increasing  interest  and  pleasure. 

In  compiling  the  present  treatise,  the  author  has  kept  this  fact  con- 
stantly in  view.  All  unnecessary  discussions  and  scholia  have  been 
avoided;  and  such  methods  have  been  adopted  as  experience  and 
attentive  observation,  combined  with  repeated  trials,  have  shown  to  be 
most  readily  comprehended.  No  attempt  has  been  made  to  render 
more  intelligible  the  simple  notions  of  position,  magnitude,  and  direc- 
tion, which  every  child  derives  from  observation  ;  but  it  is  believed 
that  these  notions  have  been  limited  and  defined  with  mathematical 
precision. 

A  few  symbols,  which  stand  for  words  and  not  for  operations,  have 
been  used,  but  these  are  of  so  great  utility  in  giving  style  and  per- 
spicuity to  the  demonstrations  that  no  apology  seems  necessary  for 
their  introduction. 

Great  pains  have  been  taken  to  make  the  page  attractive.  The 
figures  are  large  and  distinct,  and  are  placed  in  the  middle  of  the 
page,  so  that  they  fall  directly  under  the  eye  in  immediate  connec- 
tion with  the  corresponding  text.  The  given  lines  of  the  figures  are 
full  lines,  the  lines  employed  as  aids  in  the  demonstrations  are  short- 
dotted,  and  the  resulting  lines  are  long-dotted. 


IV  PREFACE. 

In  each  proposition  a  concise  statement  of  what  is  given  is  printed 
in  one  kind  of  type,  of  what  is  required  in  another,  and  the  demon- 
stration in  still  another.  The  reason  for  each  step  is  indicated  in 
email  type  between  that  step  and  the  one  following,  thus  preventing 
the  necessity  of  interrupting  the  process  of  the  argument  by  referring 
to  a  previous  section.  The  number  of  the  section,  however,  on  which 
the  reason  depends  is  placed  at  the  side  of  the  page.  The  constituent 
parts  of  the  propositions  are  carefully  marked.  Moreover,  each  distinct 
assertion  in  the  demonstrations  and  each  particular  direction  in  the 
construction  of  the  figures,  begins  a  new  line;  and  in  no  case  is  it 
necessary  to  turn  the  page  in  reading  a  demonstration. 

This  arrangement  presents  obvious  advantages.  The  pupil  perceives 
at  once  what  is  given  and  what  is  required,  readily  refers  to  the  figure 
at  every  step,  becomes  perfectly  familiar  with  the  language  of  Geom- 
etry, acquires  facility  in  simple  and  accurate  expression,  rapidly  learns 
to  reason,  and  lays  a  foundation  for  completely  establishing  the 
science. 

Original  exercises  have  been  given,  not  so  difficult  as  to  discourage 
ihe  beginner,  but  well  adapted  to  afford  an  effectual  test  of  the  degree 
in  which  he  is  mastering  the  subjects  of  his  reading.  Some  of  these 
exercises  have  been  placed  in  the  early  part  of  the  work  in  order 
that  the  student  may  discover,  at  the  outset,  that  to  commit  to  mem- 
ory a  number  of  theorems  and  to  reproduce  them  in  an  examination 
is  a  useless  and  pernicious  labor ;  but  to  learn  their  uses  and  appli- 
cations, and  to  acquire  a  readiness  in  exemplifying  their  utility  is  to 
derive  the  full  benefit  of  that  mathematical  training  which  looks  not 
so  much  to  the  attainment  of  information  as  to  the  discipline  of  the 
mental  faculties. 

G.  A.  WENTWORTH. 
Phillips  Exeter  Academy, 
1878. 


PREFACE. 


TO   THE   TEACHER. 


When  the  pupil  is  reading  each  Book  for  the  first  time,  it  will  be 
well  to  let  him  write  his  proofs  on  the  blackboard  in  his  own  lan- 
guage ;  care  being  taken  that  his  language  be  the  simplest  possible, 
that  the  arrangement  of  work  be  vertical  (without  side  work),  and 
that  the  figures  be  accurately  constructed. 

This  method  will  furnish  a  valuable  exercise  as  a  language  lesson, 
will  cultivate  the  habit  of  neat  and  orderly  arrangement  of  work, 
and  will  allow  a  brief  interval  for  deliberating  on  each  step. 

After  a  Book  has  been  read  in  this  way,  the  pupil  should  review 
the  Book,  and  should  be  required  to  draw  the  figures  free-hand.  He 
should  state  and  prove  the  propositions  orally,  using  a  pointer  to 
indicate  on  the  figure  every  line  and  angle  named.  He  should  be 
encouraged,  in  reviewing  each  Book,  to  do  the  original  exercises  ;  to 
state  the  converse  of  propositions ;  to  determine  from  the  statement, 
if  possible,  whether  the  converse  be  true  or  false,  and  if  the  converse 
be  true  to  demonstrate  it ;  and  also  to  give  well-considered  answers 
to  questions  which  may  be  asked  him  on  many  propositions. 

The  Teacher  is  strongly  advised  to  illustrate,  geometrically  and 
arithmetically,  the  principles  of  limits.  Thus  a  rectangle  with  a  con- 
stant base  6,  and  a  variable  altitude  a;,  will  afford  an  obvious  illus- 
tration of  the  axiomatic  truth  that  the  product  of  a  constant  and  a 
variable  is  also  a  variable  ;  and  that  the  limit  of  the  product  of  a 
constant  and  a  variable  is  the  product  of  the  constant  by  the  limit 
of  the  variable.  If  x  increases  and  approaches  the  altitude  a  as  a 
limit,  the  area  of  the  rectangle  increases  and  approaches  the  area  of 
the  rectangle  ab  as  a  limit;  if,  however,  x  decreases  and  approaches 
zero  as  a  limit,  the  area  of  the  rectangle  decreases  and  approaches 
zero  for  a  limit.  An  arithmetical  illustration  of  this  truth  may  be 
given  by  multiplying  a  constant  into  the  approximate  values  of  any 
repetend.  If,  for  example,  we  take  the  constant  60  and  the  repetend 
0.3333,  etc.,  the  approximate  values  of  the  repetend  will  be  ^,  -j^. 


VI  PREFACE. 

ToW^  tVAV'  6^Cm  and  these  values  multiplied  by  60  give  the  series 
18,  19.8,  19.98,  19.9998,  etc.,  which  evidently  approaches  20  as  a  limit; 
but  the  product  of  60  into  J  (the  limit  of  the  repetend  0.333,  etc.)  is 
also  20. 

Again,  if  we  multiply  60  into  the  different  values  of  the  decreasing 
series  y^,  j^,  j^jW*  Tuhns*  ®tc.,  which  approaches  zero  as  a  limit,  we 
shall  get  the  decreasing  series  2,  ^,  -^V.  sh^  etc.;  and  this  series  evi- 
dently approaches  zero  as  a  limit. 

In  this  way  the  pupil  may  easily  be  led  to  a  complete  compre- 
hension of  the  subject  of  limits. 

The  Teacher  is  likewise  advised  to  give  frequent  written  examina- 
tions. These  should  not  be  too  difficult,  and  sufficient  time  should  be 
allowed  for  accurately  constructing  the  figures,  for  choosing  the  best 
language,  and  for  determining  the  best  arrangement. 

The  time  necessary  for  the  reading  of  examination-books  will  be 
diminished  by  more  than  one-half,  if  the  use  of  the  symbols  employed 
in  this  book  be  allowed. 

G.  A.  W. 

Phillips  Exeter  Academy, 

1879. 


PREFACE.  VU 


NOTE   TO   REVISED   EDITION. 

The  first  edition  of  this  Geometry  was  issued  about  nine  years  ago. 
The  book  was  received  with  such  general  favor  that  it  has  been  neces- 
sary to  print  very  large  editions  every  year  since,  so  that  the  plates 
are  practically  worn  out.  Taking  advantage  of  the  necessity  for  new 
plates,  the  author  has  re-written  the  whole  work ;  but  has  retained 
all  the  distinguishing  characteristics  of  the  former  edition.  A  few 
changes  in  the  order  of  the  subject-matter  have  been  made,  some  of 
the  demonstrations  have  been  given  in  a  more  concise  and  simple 
form  than  before,  and  the  treatment  of  Limits  and  of  Loci  has  been 
made  as  easy  of  comprehension  as  possible. 

More  than  seven  hundred  exercises  have  been  introduced  into  this 
edition.  These  exercises  consist  of  theorems,  loci,  problems  of  con- 
struction, and  problems  of  computation,  carefully  graded  and  specially 
adapted  to  beginners.  Nq_geometry  can  now  receive  favor  unless  it 
provides  exercises  for  independent  investigation,  which  must  be  of  such 
a  kind  as  to  interest  the  student  as  soon  as  he  becomes  acquainted 
with  the  methods  and  the  spirit  of  geometrical  reasoning.  The  author 
has  observed  with  the  greatest  satisfaction  the  rapid  growth  of  the 
demand  for  original  exercises,  and  he  invites  particular  attention  to 
the  systematic  and  progressive  series  of  exercises  in  this  edition. 

The  part  on  Solid  Geometry  has  been  treated  with  much  greater 
freedom  than  before,  and  the  formal  statement  of  the  reasons  for  the 
separate  steps  has  been  in  general  omitted,  for  the  purpose  of  giving  a 
more  elegant  form  to  the  demonstrations. 

A  brief  treatise  on  Conic  Sections  (Book  IX)  has  been  prepared, 
and  is  issued  in  pamphlet  form,  at  a  very  low  price.  It  will  also  be 
bound  with  the  Geometry  if  that  arrangement  is  found  to  be  gen- 
erally desired. 


Vlll  PREFACE. 

The  author  takes  this  opportunity  to  express  his  grateful  appre- 
ciation of  the  generous  reception  given  to  the  Geometry  heretofore  by 
the  great  body  of  teachers  throughout  the  country,  and  he  confidently 
anticipates  the  same  generous  judgment  of  his  efforts  to  bring  the  work 
up  to  the  standard  required  by  the  great  advance  of  late  in  the  sci- 
ence and  method  of  teaching. 

The  author  is  indebted  to  many  correspondents  for  valuable  sug- 
gestions ;  and  a  special  acknowledgment  is  due,  for  criticisms  and 
careful  reading  of  proofs,  to  Messrs.  C.  H.  Judson,  of  Greenville,  S.G. ; 
Samuel  Hart,  of  Hartford,  Conn. ;  J.  M.  Taylor,  of  Hamilton,  N.Y. ; 
W.  Le  Conte  Stevens,  of  Brooklyn,  N.Y. ;  E.  R.  Offutt,  of  St.  Louis, 
Mo. ;  J.  L.  Patterson,  of  Lawrenceville,  'N.J. ;  G.  A.  Hill,  of  Cam- 
bridge, Mass. ;  T.  M.  Blaksiee,  Des  Moines,  la. ;  G.  W.  Sawin,  of  Cam- 
bridge, Mass. ;  and  Ira  M.  De  Long,  of  Boulder,  Col. 

Corrections  or  suggestions  will  be  thankfully  received. 

G.  A.  WENTWORTH 
Phillips  Exetee  Academy, 
1888. 


COl^TENTS. 


GEOMETRY. 

PACK 

Definitioits 1 

Straight  Lines 5 

Plane  Angles 7 

Magnitude  of  Angles 9 

Angular  Units 10 

Method  of  Superposition 11 

Symmetry 13 

Mathematical  Terms .  14 

Postulates 15 

Axioms 16 

Symbols 16 


PLANE  GEOMETRY. 

BOOK  I.    The  Straight  Line. 

The  Straight  Line 17 

Parallel  Lines 22 

Perpendicular  and  Oblique  Lines             ....  33 

Trianqlf^ 40 

Quadrilaterals 66 

Polygons  in  General    ........  66 

Exercises 72 


X  CONTENTS. 

BOOK  II.     The  Circle. 

•  PAOB 

Definitions .  75 

Arcs  and  Choeds 77 

Tangents         .        .        .^ 89 

Measueement 92 

Theoey  of  Limits 94 

Measure  of  Angles 98. 

Problems  of  Construction 106 

Exercises 126 

BOOK  III.   Proportional  Lines  and  Similar  Polygons. 

Theory  of  Proportion '^ .        .  131 

Proportional  Lines 138 

Similar  Triangles 145 

Similar  Polygons 153 

Numerical  Properties  of  Figures 156 

Problems  op  Construction 167 

Problems  of  Computation  ''.        .        .        .        .        ,        .173 

Exercises ''.  176 

BOOK  IV.    Areas  of  Polygons. 

Areas  of  Polygons        ...'.....  181 

Comparison  of  Polygons .  188 

Problems  of  Construction 19i^ 

Problems  of  Computation     .        , 204 

Exercises 205 

BOOK  V.     Regular  Polygons  and  Circles. 

Regular  Polygons  and  Circles  ,               .        ...        .  209 

Peoblems  of  Construction    .        .    •    .       ,       .       .        .  222 

Maxima  and  Minima            •  230 

Exercises 237 

Miscellaneous  Exercises 240 


CONTENTS.  ■  XI 


SOLID   GEOMETKY. 
BOOK  VI.     Lines  and  Planes  in  Space. 

PAGE 

Definitions 243 

Perpendicular  Lines  and  Planes 246 

Parallel  Lines  and  Planes 253 

Dihedral  Angles 260 

Measure  of  Dihedral  Angles     .        .        .        .        .        .  262 

Planes  Perpendicular  to  Each  Other       ....  265 

Angle  of  a  Straight  Line  and  a  Plane  ....  270 

A  Perpendicular  Between  Two  Straight  Links              .  271 

Polyhedral  Angles 272 

Symmetrical  Polyhedral  Angles 273 

BOOK  VII.     Polyhedrons,  Cylinders,  and  Cones. 

Prisms  and  Parallelopipeds 279 

Pyramids 294 

Similar  Polyhedrons 308 

Regular  Polyhedrons 312 

General  Theorems  of  Polyhedrons 317 

Cylinders 319 

Cones 325 

Numerical  Exercises 334 


BOOK  VIII.     The  Sphere. 

Plane  Sections  and  Tangent  Planes  . 
Figures  on  the  Surface  of  a  Sphere 
Measuremei^t  of  Spherical  Surfaces. 
Volume  of  a>ySphere     . 
Numerical  Exercises     .... 

MiSCELLANECUr   J^ERCISES 


339 
350 
366 
375 
380 
383 


GEOMETRY. 


DEFINITIONS. 

1.  If  a  block  of  wood  or  stone  be  cut  in  the  shape  repre 
sented  in  Fig.  1,  it  will  have  six  flat  faces. 

Each  face  of  the  block  is  called 
a  surface;  and  if  these  faces  are  made  ^ 
smooth  by  polishing,  so  that,  when 
a  straight-edge  is  applied  to  any  one 
of  them,  the  straight  edge  in  every 
part  will  touch  the  surface,  the  faces 
2i,VQ  coWedi  plane  surfaces^  ov  planes. 

2.  The  edge  in  which  any  two  of  these  surfaces  meet  is 
called  a  line. 

3.  The  corner  at  which  any  three  of  these  lines  meet  is 
called  a  point. 

4.  For  computing  its  volume,  the  block  is  measured  in  three 
principal  directions : 

From  left  to  right,  A  to  B. 
From  front  to  back,  A  to  C. 
From  bottom  to  top,  A  to  D. 

These  three  measurements  are  called  the  dimensions  of  the 
block,  and  are  named  length,  breadth  (or  width),  thickness 
{height  or  depth). 


2  ^  GEOMETRY. 

A  solid,  therefore,  has  three  dimensions,  length,  breadth,  and 
thickness. 

6.  The  surface  of  a  solid  is  no  part  of  the  solid.  It  is 
simply  the  boundary  or  limit  of  the  solid.  A  surface,  there- 
fore, has  only  tivo  dimensions,  length  and  breadth.  So  that, 
if  any  number  of  flat  surfaces  be  put  together,  they  will  coin- 
cide and  form  one  surface. 

6.  A  line  is  no  part  of  a  surface.  It  is  simply  a  boundary 
or  limit  of  the  surface.  A  line,  therefore,  has  only  one  dimen- 
sion, length.  So  that,  if  any  number  of  straight  lines  be  put 
together,  they  will  coincide  and  form  one  line. 

7.  A  point  is  no  part  of  a  line.  It  is  simply  the  limit  of 
the  line.  A  point,  therefore,  has  no  dimension,  but  denotes 
position  simply.  So  that,  if  any  number  of  points  be  put 
together,  they  will  coincide  and  form  a  single  point. 

8.  A  solid,  in  common  language,  is  a  limited  portion  of 
STp&ce  filled  with  matter ;  but  in  Geometry  we  have  nothing  to 
do  with  the  matter  of  which  a  body  is  composed ;  we  study 
simply  its  shape  and  size;  that  is,  we  regard  a  solid  as  a  lim- 
ited portion  of  space  which  may  be  occupied  by  a  physical 
body,  or  marked  out  in  some  other  way.     Hence, 

A  geometrical  solid  is  a  limited  portion  of  space. 

9.  It  must  be  distinctly  understood  at  the  outset  that  the 
points,  lines,  surfaces,  and  solids  of  Geometry  are  purely  ideal, 
though  they  can  be  represented  to  the  eye  in  only  a  material 
way.  Lines,  for  example,  drawn  on  paper  or  on  the  black- 
board, will  have  some  width  and  some  thickness,  and  will  so 
far  fail  of  being  true  lines ;  yet,  when  they  are  used  to  help 
the  mind  in  reasoning,  it  is  assumed  that  they  represent  per- 
fect lines,  without  breadth  and  without  thickness. 


DEFINITIONS.  3 

10.  A  point  is  represented  to  the  eye  by  a  fine  dot,  and 
named  by  a  letter,  as  A  (Fig.  2) ;  a  line  is  named  by  two 
letters,  placed  one  at  each  end, 
as  £F;  a  surface  is  represented 
and  named  by  the  lines  which 
bound  it,  as  BCDF\  a  solid  is 
represented  by  the  faces  which 
bound  it.  Fig.  2. 

11.  By  supposing  a  solid  to  diminish  gradually  until  it 
vanishes  we  may  consider  the  vanishing  point,  a  jpomi  in  space, 
independent  of  a  line,  having  position  but  no  extent, 

12.  If  a  point  moves  continuously  in  space,  its  path  is  a  line. 
This  line  may  be  supposed  to  }M  of  unlimited  extent,  and  may 
be  considered  independent  of  the  idea  of  a  surface. 

13.  A  surface  may  be  conceived  as  generated  by  a  line 
moving  in  space,  and  as  of  unlimited  extent,  A  surface  can 
then  be  considered  independent  of  the  idea  of  a  solid. 

14.  A  solid  may  be  conceived  as  generated  by  a  surface  in 
motion. 

Thus,  in  the  diagram,  let  the  up-  d 

right  surface   ABCD  move  to  the  / 

nght  to  the  position  EFOH.     The  "^X"" 

points  A,  B,  C,  and  D  will  generate  [c 

the  lines  AE,  BF,  CO,   and   DH,  ^"""  l^Il^ 

respectively.   '  The   lines  AB,  BC,  Fig.  3. 
CD,  and  AD  will  generate  the  sur- 
faces  AF,  BO,   CH,   and   AH,   respectively.     The   surface 
ABCD  will  generate  the  solid  AO. 

15.  Geometry  is  the  science  which  treats  of  position,  form, 
and  magnitude. 

16.  Points,  lines,  surfaces,  and  solids,  with  tneir  relations, 
constitute  the  subject-matter  of  Geometry. 


4  GEOMETRY. 

17.  A  straight  line,  or  right  line,  is  a  line  which  has  the 

same  direction  throughout  its      * g 

whole  extent,  as  the  line  AB. 

18.  A  curved  line  is  a  line 
no  part  of  which  is  straight, 
as  the  line  CD. 

19.  A  broken  line  is  a  series 
of  different  successive  straight 
lines,  as  the  line  EF. 

20.  A  mixed  line  is  a  line  composed  of  straight  and  curved 
lines,  as  the  line  GS. 

A  straight  line  is  often  called  simply  a  line,  and  a  curved 
line,  a  curve.  * 

21.  A  plane  surface,  or  a  plane,  is  a  surface  in  which,  if 
any  two  points  be  taken,  the  straight  line  joining  these  points 
will  lie  wholly  in  the  surface. 

22.  A  curved  surface  is  a  surface  no  part  of  which  is  plane. 

23.  Figure  or  form  depends  upon  the  relative  position  of 
points.  Thus,  the  figure  or  form  of  a  line  (straight  or  curved) 
depends  upon  the  relative  position  of  the  points  in  that  line  ; 
the  figure  or  form  of  a  surface  depends  upon  the  relative  posi- 
tion of  the  points  in  that  surface. 

24.  With  reference  to  form  or  shape,  lines,  surfaces,  and 
solids  are  colled,  figures. 

With  reference  to  extent,  lines,  surfaces,  and  solids  are 
called  magnitudes.  » 

25.  A  plane  figure  is  a  figure  all  points  of  which  are  in  the 
same  plane. 

26.  Plane  figures  formed  by  straight  lines  are  called  rec- 
tilinear figures;  those  formed  by  curved  lines  are  called 
curvilinear  figures  ;  and  those  formed  by  straight  and  curved 
lines  are  called  mixtilinear  figures. 


DEFINITIONS.  O 

27.  Figures  which  have  the  same  shape  are  called  sirmlar 

figures.  Figures  which  have  the  same  size  are  called  equiva- 
lent figures.  Figures  which  have  the  same  shape  and  size 
are  called  equal  or  congruent  figures. 

28.  Geometry  is  divided  in  two  parts,  Plane  Geometry  and 
Solid  Geometry.  Plane  Geometry  treats  of  figures  all  points 
of  which  are  in  the  same  plane.  Solid  Geometry  treats  of 
figures  all  points  of  which  are  not  in  the  same  plane. 


Straight  Lines. 

29.  Through  a  point  an  indefinite  number  of  straight  lines 
may  be  drawn.     These  lines  will  have  difierent  drections. 

30.  If  the  direction  of  a  straight  line  and  a  point  in  the 
line  are  known,  the  position  of  the  line  is  known  ;  in  other 
words,  a  straight  line  is  determined  if  its  direction  and  one  of 
its  points  are  known.     Hence, 

All  straight  lines  which  pass  through  the  same  point  in  the 
same  direction  coincide,  and  form  hut  one  line. 

31.  Between  two  points  one,  and  only  one,  straight  line  can 
be  drawn  ;  in  other  words,  a  straight  line  is  determined  if  tw<? 
of  its  points  ara  known.     Hence, 

Two  straight  lines  which  have  two  points  coTumon  coincidi 
throughout  their  whole  extent,  andjorm'hut  one  liiie. 

32.  Two  straight  lines  can  intersect  (cut  each  other)  in  only 
one  point ;  for  if  they  had  two  points  common,  they  would 
coincide  and  not  intersect. 

33.  Of  all  lines  joining  two  points  the  shortest  is  the  straight 
line,  and  the  length  of  the  straight  line  is  called  the  distance 
between  the  two  points. 


O  GEOMETRY, 

84.  A  straight  line  determined  by  two  points  is  considered 
as  prolonged  indefinitely  both  ways.  Such  a  line  is  called  an 
indefinite  straight  line. 

35.  Often  only  the  part  of  the  line  between  two  fixed  points 
is  considered.     This  part  is  then  called  a  segment  of  the  line. 

For  brevity,  we  say  "the  line  AB''  to  designate  a  segment 
of  a  line  limited  by  the  points  A  and  B, 

36.  Sometimes,  also,  a  line  is  considered  as  proceeding  from 
a  fixed  point  and  extending  in  only  one  direction.  This  fixed 
point  is  then  called  the  origin  of  the  line. 

37.  If  any  point  Cbe  taken  in  a  given  straight  line  AB^  the 
two  parts  CA  and  CB  are 

said  to  havr  opposite  direc-     ^ ^ ^ 

tions  from  the  point  C.  Fig.  5. 

38.  Every  straight  line,  as  AB,  may  be  considered  as  hav- 
ing opposite  directions,  namely,  from  A  towards  B,  which  is 
expressed  by  saying  "line  AB''\  and  from  B  towards  A,  which 
is  expressed  by  saying  "line  BA.'* 

39.  If  the  magnitude  of  a  given  line  is  changed,  it  becomes 
longer  or  shorter. 

Thus  (Fig.  5),  by  prolonging  AQ  to  B  we  add  OB  to  AC, 
and  AB  =  AQ-{-  CB.  By  diminishing  AB  to  (7,  we  subtract 
CB  from  AB,  and  AC==AB-  CB. 

If  a  given  line  increases  so  that  it  is  prolonged  by  its  own 

magnitude  several  times  in       .  ^  ^  ^^  7-r 

°.         ,,.       .  ,.-4  B  G  D  E 

succession,  the  line  is  multi-     h 1 1 ' •- 

plied,  and  the  resulting  line  ^^'   ' 

is  called  a  multiple  of  the  given   line.      Thus   (Fig.   6),   if 

AB  =^  BC=  CD^  DE,  i^iQXi  AC=^2AB,  AI)  =  SAB,  and 

AU=^AB.   Aho,AB^iAC,AB^iAD,d.ndAB=kAE. 

Hence, 


DEFINITIONS.  "^--^^  7 

Lines  of  given  length  may  he  added  and  subtracted;   they 
may  also  he  multiplied  and  divided  hy  a  number. 


Fig.  7. 


Plane  Angles. 

40.  The  opening  between  two  straight  lines  which  meet  is 
called  a  plane  angle.  The  two  lines  are  called  the  sides,  and 
the  point  of  meeting,  the  vertex,  of  the  angle. 


41.  If  there  is  but  one  angle  at  a 
given  vertex,  it  is  designated  by  a  cap- 
ital letter  placed  at  the  vertex,  and  is 
read  by  simply  naming  the  letter ;  as, 
angle  A  (Fig.  7). 

But  when  two  or  more  angles  have 
the  same  vertex,  each  angle  is  desig- 
nated by  three  letters,  as  shown  in 
Fig.  8,  and  is  read  by  naming  the 
three  letters,  the  one  at  the  vertex  be- 
tween the  others.  Thus,  the  angle 
DAO  means  the  angle  formed  by  the 
sides  AD  and  AC. 

It  is  often  convenient  to  designate 
an  angle  by  placing  a  small  italic  let- 
ter between  the  sides  and  near  the 
vertex,  as  in  Fig.  9. 

42.  Two  angles  are  equal  if  they 
can  be  made  to  coincide. 


FiQ.  9. 


43.  If  the  line  AD  (Fig.  8)  is  drawn  so  as  to  divide  the 
angle  BAC  into  two  equal  parts,  BAD  and  CAD,  AD  is 
called  the  bisector  of  the  angle  BAC  In  general,  a  line  that 
divides  a  geometrical  magnitude  into  two  equal  parts  is  called 
a  bisector  of  it. 


8 


GEOMETRY, 


44.  Two  angles  are  called  ad- 
jacent when  they  have  the  same 
vertex  and  a  common  side  be- 
tween them ;  as,  the  angles  J50D 
and  AOB  (Fig.  10). 

45.  When  one  straight  line 
stands  upon  another  straight  line 
and  makes  the  adjacent  angles 
equal,  each  of  these  angles  is 
called  a  right  angle.  Thus,  the 
equal  angles  DCA  and  DCB 
(Fig.  11)  are  each  a  right  angle. 


0 

Fig.  10. 


C 
Fig.  11. 


—B 


46.  When  the  sides  of  an  an- 
gle extend  in  opposite  directions, 

so  as  to  be  in  the  same  straight  line,  the  angle  is  called  a 
straight  angle.  Thus,  the  angle  formed  at  C(Fig.  11)  with 
its  sides  CA  and  CB  extending  in  opposite  directions  from  C, 
is  a  straight  angle.  Hence  a  right  angle  may  be  defined  as 
half  a  straight  angle, 

47.  A  perpendicular  to  a  straight  line  is  a  straight  line  that 
makes  a  right  angle  with  it.  Thus,  if  the  angle  DCA  (Fig.  11^ 
is  a  right  angle,  DO  is  perpendicular  to  AB,  and  AB  is  per- 
pendicular to  DC. 

48.  The  point  (as  C,  Fig.  11)  where  a  perpendicular  me«jt8 
another  line  is  called  the  foot  of  the  perpendicular. 

49.  Every  angle  less  than  a  right  an- 
gle is  called  an  acute  angle;  as,  angle  A. 

Fig    12 
60.  Every  angle  greater  than  a  right 

angle  and  less  than  a  straight  angle  is  called  an  obtuse  angle; 

as,  angle  C(Fig.  13). 


DEFINITIONS. 


9 


51.  Every  angle  greater  than  a  straight  angle  and  less 
than  two  straight  angles  is  called  a  reflex  angle;  as,  angle  0 
(Fig.  14). 


KlG.    1.^. 


Fig.  14. 


52.  Acute,  obtuse,  and  reflex  angles,  in  distinction  from 
right  and  straight  angles,  are  called  oblique  angles ;  and  inter- 
secting lines  that  are  not  perpendicular  to  each  other  are 
called  oblique 


53.  When  two  angles  have  the  same  vertex,  and  the  sides 
of  the  one  are  prolongations  of 
the  sides  of  the  other,  they  are 
called  vertical  angles.  Thus,  a 
and  b  (Fig.  15)  are  vertical  an- 
gles. 


FiQ.  15. 


54.  Two   angles    are    called 
complementary  when  their  sum 

is  equal  to  a  right  angle ;  and  each  is  called  the  complement 
of  the  other;  as,  angles  DOB  and  DOC  (Fig.  10). 

55.  Two  angles  are  called  supplementary  when  their  sum  is 
equal  to  a  straight  angle ;  and  each  is  called  the  supplement 
of  the  other;  as,  angles  DOB  and  DO  A  (Fig.  10). 


Magnitude  of  Angles. 

56.   The  size  of  an  angle  depends  upon  the  extent  of  opening 
of  its  sides,  and  not  upon  their  length.     Suppose  the  straight 


10 


GEOMETRY. 


line  OQ  to  move  in  the  plane  of  the  paper  from  coincidence 
with  OA,  about  the  point  0  as  a  pivot,  to  the  position  00\ 
then  the  line  00  describes  or  generates 
the  angle  A  00. 

The  amount  of  rotation  of  the  line 
from  the  position  OA  to  the  position 
OCis  the  acute  angle  AOO. 

If  the  rotating  line  moves  from  the 
position  OA  to  the  position  0£,  perpen- 
dicular to  OA,  it  generates  the  right 
angle  AOB ;   if  it  moves  to  the  position 

CD,  it  generates  the  obtuse  angle  AOD ;  if  it  moves  to  the  posi- 
tion OA^,  it  generates  the  straight  angle  AOA' ;  if  it  moves  to 
the  position  OB',  it  generates  the  reflex  angle  AOB\  indicated 
by  the  dotted  line ;  and  if  it  continues  its  rotation  to  the  posi- 
tion OA,  whence  it  started,  it  generates  two  straight  angles. 

Hence  the  whole  angular  magnitude  about  a  point  in  a 
plane  is  equal  to  two  straight  angles,  or  four  right  angles;  and 
the  angular  magnitude  about  a  point  on  one  side  of  a  straight 
line  drawn  through  that  point  is  equal  to  one  straight  angle, 
or  two  right  angles. 

Angles  are  magnitudes  that  can  be  added  and  subtracted ; 
they  may  also  be  multiplied  and  divided  by  a  number. 


AiwjuLAR  Units. 

57.  If  we  suppose  00  (Fig.  17)  to 
turn  about  0  from  a  position  coinci- 
dent with  OA  until  it  makes  a  com- 
plete revolution  and  comes  again  into 
coincidence  with  OA,  it  will  describe 
the  whole  angular  magnitude  about 
the  -point  0,  while  its  end  point  0 
will  describe  a  curve  called  a  circuw,- 
ference. 


DEFINITIONS.  11 

58.  By  adopting  a  suitable  unit  of  angles  we  are  able  to 
express  the  magnitudes  of  angles  in  numbers. 

If  we  suppose  00  (Fig.  17)  to  turn  about  0  from  coinci- 
dence with  OA  until  it  makes  one  three  hundred  and  sixtieth 
of  a  revolution,  it  generates  an  angle  at  0,  which  is  taken 
as  the  unit  for  measuring  angles.  This  unit  is  called  a 
degree. 

The  degree  is  subdivided  into  sixty  equal  parts  called 
minutes,  and  the  minute  into  sixty  equal  parts,  called  seconds. 

Degrees,  minutes,  and  seconds  are  denoted  by  symbols. 
Thus,  5  degrees  13  minutes  12  seconds  is  written,  5°  13'  12". 

A  right  angle  is  generated  when  00  has  made  one-fourth 
of  a  revolution  and  is  an  angle  of  90**;  a  straight  angle  is 
generated  when  00  has  made  one-half  of  a  revolution  and 
is  an  angle  of  180° ;  and  the  whole  angular  magnitude  about 
0  IS  generated  when  00  has  made  a  complete  revolution,  and 
contains  360°. 

The  natural  angular  unit  is  one  complete  revolution.  But 
the  adoption  of  this  unit  would  require  us  to  express  the 
values  of  all  angles  by  fractions.  The  advantage  of  using  the 
degree  as  the  unit  consists  in  its  convenient  size,  and  in  the  fact 
that  360  is  divisible  by  so  many  different  integral  numbers. 


Method  of  Superposition. 

59.  The  test  of  the  equality  of  two  geometrical  magnitudes 
IS  that  they  coincide  throughout  their  whole  extent. 

Thus,  two  straight  lines  are  equal,  if  they  can  be  so  placed 
that  the  points  at  their  extremities  coincide.  Two  angles  are 
equal,  if  they  can  be  so  placed  that  they  coincide. 

In  applying  this  test  of  equality,  we  assume  that  a  line  may 
be  moved  from  one  place  to  another  without  altering  its  length; 
that  an  angle  may  be  taken  up,  turned  over,  and  put  down, 
without  altering  the  difference  in  direction  of  its  sides. 


12 


GEOMETRY. 


Fig.  18. 

This  method  enables  us  to  compare  magnitudes  of  the  same 
kind.  Suppose  we  have  two  angles,  ABC  and  DEF.  Let 
the  side  ED  be  placed  on  the  side  BA,  so  that  the  vertex  E 
shall  fall  on  B ;  then,  if  the  side  EF  falls  on  BC,  the  angle 
DEF  equals  the  angle  ABQ\  if  the  side  EF  falls  between 
-SCand  BA  in  the  direction  BG,  the  angle  DEF\%  less  than 
ABO\  but  if  the  side  ^i^  falls  in  the  direction  BH,  the  angle 
DEF  ia  greater  than  ABO. 

This  method  enables  us  to  add  magnitudes  of  the  same  kind. 
Thus,  if  we  have  two  straight  lines      BC 


AB  and  CD,  by  placing  the  point    ^ ^ 

0  on  B,  and  keeping  CD  in  the   ^ ^ 

same  direction  with  AB,  we  shall  Fig.  19. 

have  one  continuous  straight  line  AD  equal  to  the  sum  of 

the  lines  ^5  and  CD. 

C 

/F 


b:^ 


Fig.  20. 


B 

Fig.  21. 


Again  :  if  we  have  the  angles  ABC  and  DEF,  and  place 
the  vertex  E  on  B  and  the  side  ED  in  the  direction  of  BC,  the 
angle  DEF  will  take  the  position  CBIT,  and  the  angles  DEF" 
and  ABC  will  together  equal  the  angle  ABE". 

If  the  vertex  E  is  placed  on  B,  and  the  side  ED  on  BA,  the 
angle  i).£'i^will  take  the  position  ABF,  and  the  angle  FBO 
will  be  the  difference  between  the  angles  ABC  and  DEF. 


DEFINITIONS. 


13 


Symmetry. 

60.   Two  points  are  said  to  be  symmeti-ical  with  respect  to  a 
third  point,  called  the  centre  of  sym-  ^ 

metry,  if  this  third  point  bisects  the    P'— 
straight  line  which  joins  them.  Thus, 


— » 

Fig.  22. 


Pand  P'  are  symmetrical  with  respect  to  (7  as 
bisects  the  straight  line  PP\ 

61.  Two  points  are  said  to  be  sym- 
metrical  with  respect  to  a  straight 
line,  called  the  axis  of  symmetry,  if 

this   straight    line   bisects   at    right        X- 

angles  the  straight  line  which  joins 
them.  Thus,  P  and  P'  are  symmet- 
rical with  respect  to  XX*  as  an  axis, 
if  XX*  bisects  PP*  at  right  angles. 

62.  Two  figures  are  said  to  be  sym- 
metrical with  respect  to  a  centre  or 
an  axis  if  every  point  of  one  has  a 
corresponding  symmetrical  point  in 
the  other. '  Thus,  if  every  point  in 
the  figure  A*B^C*  has  a  symmetrical 
point  in  ABC,  with  respect  to  D  as 
a  centre,  the  figure  A*B*C*  is  sym- 
metrical to  ABQ  with  respect  to  D 
as  a  centre. 

63.  If  every  point  in  the  figure 
A'B'C  has  a  symmetrical  point  in 
ABO,  with  respect  to  XX*  as  an 
axis,  the  figure  A*B'0*  is  symmetri- 
cal to  ABC  with  respect  to  XX*  as 
an  axis. 


centre,  if  C 


P' 

Fig.  23. 


14 


GEOMETRY. 


64.  A  figure  is  symmetrical  with  re- 
spect to  a  point,  if  the  point  bisects 
every  straight  line  drawn  through  it 
and  terminated  by  the  boundary  of  the 
figure. 

65.  A  plane  figure  is  symmetrical  with 
respect  to  a  straight  line,  if  the  line 
divides  it  into  two  parts,  which  are  sym- 
metrical with  respect  to  this  straight 
line. 

Mathematical  Terms. 


Fia.  27. 


66.    A  "proof  or  demonstration  is  a  course  of  reasoning  by 
which  the  truth   or  falsity   of  any   statement  is    logically 

established. 

\ 


67.   A  theorem  is  a  statement  to  be  proved. 


68.  A  theorem  consists  of  two  parts:  the  hypothesis,  or 
that  which  is  assumed ;  and  the  conclusion,  or  that  which  is 
asserted  to  follow  from  the  hypothesis. 

69.  An  axiom  is  a  statement  the  truth  of  which  is  admitted 
without  proof. 

70.  A  construction  is  a  graphical  representation  of  a  geo- 
metrical figure, 

71.  A  'problem  is  a  question  to  be  solved. 

72.  The  solution  of  a  problem  consists  of  four  parts  : 

(1)  The  analysis,  or  course  of  thought  by  which  the  con- 
struction of  the  required  figure  is  discovered ; 

(2)  The  construction  of  the  figure  with  the  aid  of  ruler  and 
compasses ; 

(3)  The  proof  that  the  figure  satisfies  all  the  given  condi- 
tions: 


DEFINITIONS.  15 

(4)  The  discussion  of  the  limitations,  which  often  exist, 
within  which  the  solution  is  possible. 

73.  A  postulate  is  a  construction  admitted  to  be  possible. 

74.  A  proposition  is  a  general  term  for  either  a  theorem 
or  a  problem. 

75.  A  coi'ollary  is  a  truth  easily  deduced  from  the  propo- 
sition to  which  it  is  attached.   • 

76.  A  scholium  is  a  remark  upon  some  particular  feature 
of  a  proposition. 

77.  The  converse  of  a  theorem  is  formed  by  interchanging 
its  hypothesis  and  conclusion.     Thus, 

If  A  is  equal  to  B,  C  is  equal  to  D.   (Direct.) 
If  O  is  equal  to  D,  A  is  equal  to  B.   (Converse.) 

78.  The  opposite  of  a  proposition  is  formed  by  stating  the 
negative  of  its  hypothesis  and  its  conclusion.     Thus, 

If  A  is  equal  to  B,  C  is  equal  to  D.    (Direct.) 

If  A  is  not  equal  to  B,  Q  is  not  equal  to  D.   (Opposite.) 

79.  The  converse  of  a  truth  is  not  necessarily  true.  Thus, 
Every  horse  is  a  quadruped  is  a  true  proposition,  but  the  con- 
verse, Every  quadruped  is  a  horse,  is  not  true. 

^,  If  a  direct  proposition  and  its  converse  are  true,  the  op- 
posite proposition  is  true;  and  if  a  direct  proposition  and  its 
opposite  are  trice,  the  converse  proposition  is  true. 

81.  Postulates. 

Let  it  be  granted  — 

1.  That  a  straight  line  can  be  drawn  from  any  one  point  to 
any  other  point. 

2.  That  a  straight  line  can  be  produced  to  any  distance,  or 
can  be  terminated  at  any  point. 

3.^  That  a  circumference  may  be  described  about  any  point 
as  a  centre  with  a  radius  of  given  length. 


16  GEOMETRY. 


Axioms. 


*^  1.   Things  which  are  equal  to  the  same  thing  are  equal  to 

each  other. 

I''    2.    If  equals  are  added  to  equals  the  sums  are  equal. 
^,    3.    If  equals  are  taken  from   equals   the   remainders   are 

equal. 

4.  If  equals  are  added  to  unequals  the  sums  are  unequal, 
and  the  greater  sum  is  obtained  from  the  greater  magnitude. 

5.  If  equals  are  taken  from  unequals  the  remainders  are 
unequal,  and  the  greater  remainder  is  obtained  from  the 
greater  magnitude. 

6.  Things  which  are  double  the  same  thing,  or  equal  things, 
are  equal  to  each  other. 

7.  Things  which  are  halves  of  the  same  thing,  or  of  equal 
things,  are  equal  to  each  other. 

8.  The  whole  is  greater  than  any  of  its  parts. 

9.  The  whole  is  equal  to  all  its  parts  taken  together. 

83.  Symbols  and  Abbreviations. 

+  increased  by,  O  circle.    ©  circles. 

—  diminished  by.  Def. definition. 

X  multiplied  by.  '  Ax axiom. 

-*-  divided  by.  Hyp-  •  •  •  hypothesis. 

=  is  (or  are)  equal  to:  Cor corollary^ 

o  is  (or  are)  equivalent  to.  Adj adjacent. 

>  is  (or  are)  greater  than,  Iden. . . .  identical. 

<  is  (or  are)  less  than.  Cons, . . .  construction. 

.*.  therefore.  Sup supplementary. 

Z  angle.  Sup.-adj.  supplementary-adjacent. 

A  angles.  Ext.-int.  exterior-interior. 

X  perpendicular.  Alt.-int.   alternate-interior. 

Ji  perpendiculars.  Ex exercise. 

II  parallel. .  rt right. 

lis  parallels.  st straight. 

A  triangle.  q.e.d quod  erat  demonstrandum, 

A  triangles.  which  was  to  he  proved 

O  parallelogram.  q.e.p.  . . .  quod  erat  faciendum,  v)hich 
[EJ  parallelograms.  was  to  he  done. 


PLAIN^E  GEOMETRY 

BOOK   I. 

THE    STRAIGHT    LINE. 

^  

Proposition  I.    Theorem. 
84.  Ml  straight  angles  are  equal, 

F 


A ^ B 

E 


Let  /.EC A  and  Z.  FED  he  any  two  straight  angles. 

To pT(yve  ABCA^A  FED. 

Proof.   Apply  the  ABCA\.q  the  A  FED,  so  that  the  vertex 
C  shall  fall  on  the  vertex  E,  and  the  side  CB  on  the  side  EF. 

Then  CA  will  coincide  with  ED^ 
(5ecaws6  BCA  and  FED  are  straight  lines  and  have  two  points  common). 

.     Therefore  the  Z  BOA  is  equal  to  the  Z  FED.  §  59 

85.  Cor.  1.   All  nght  angles  are  equal. 

86.  Cor.  2.   The  angular  units,  degree,  minute,  and  second, 
have  constant  values. 

87.  Cor.  3.   The  complements  of  equal  angles  are  eqvxxl. 

88.  Cor.  4.   The  supplements  of  equal  angles  are  equal. 

89.  Cor.  5.  At  a  given  point  in  a  given  straight  line  one 
perpendicular,  and  only  one,  can  he  erected. 


18  PLANE   GEOMETRY.  —  BOOK   I. 


Proposition  II.     Theorem. 

90.  If  two  adjacent  angles  have  their  exterior  sides 
in  a  straigM  line,  these  angles  are  supplements  of 
each  other. 


Let  the  exterior  sides  OA  and  OB  of  the  adjacent 
AAOD  and  BOD  be  in  the  straight  line  AB. 

To  prove  A  AOD  and  BOD  supplementary. 

Proof.                    AOB  is  a  straight  line.  Hyp. 

.-.  the  Z^O^  is  a  St.  Z.  §46 

But         the  AAOB+  BOB  =  the  st.  Z  AOB.  Ax.  9 

.*,  the  A  AOB  and  BOB  are  supplementary.        §  55 

aE.  D. 

91.  Scholium.  Adjacent  angles  that  are  supplements  of 
each  other  are  called  supplementary-adjacent  angles. 

92.  Cor.  Since  the  angular  magnitude  about  a  point  is 
neither  increased  nor  diminished  by  the  number  of  lines  which 
radiate  from  the  point,  it  follows  that, 

ITie  sum  of  all  the  angles  about  a  point  in  a  plane  is  equal 
to  two  straight  angles,  or  four  right  angles. 

The  sum  of  all  the  angles  about  a  point  on  the  same  side  of  a 
straight  line  passing  through  the  point,  is  equal  to  a  straight 
angle,  or  two  right  angles. 


THE   STRAIGHT   LINE.  19 


Peoposition  III.    Theorem. 

93.  CoNVEBSELY  :  If  two  adjoceiit  angles  are  supple^ 
merits  of  en/^h  other,  their  exterior  sides  lie  in  the 
^ame  straight  line. 


AC  B  F 

Let  the  adjacent  A  OCA  +  OCB  =  a  straight  angle.     ^ 
To  prove  A  C  and  CB  in  the  same  straight  line. 
Proof.    Suppose  CF  to  be  in  the  same  line  with  AC.       §  81 
Then         /.  OCA  +  Z.  OCF  is  a  straight  angle.  §  90 

But  Z.  OCA  +  Z  OCB  is  a  straight  angle.  Hyp. 

.-.  Z  OCA  +  Z  OCF=^  Z  OCA  +  Z  OCB.       Ax.  1 
Take  away  from  each  of  these  equals  the  common  Z  OCA. 
Then  Z  OCF=  Z  OCB,  Ax.  3 

.*.  CB  and  <?i^  coincide. 
.'.  -4  (7  and  CB  are  in  the  same  straight  line.       q.  e.  d. 

94.  Scholium.  Since  Propositions  II.  and  III.  are  true, 
their  opposites  are  true ;  namely,  §  80 

If  the  exterior  sides  of  two  adjacent  angles  are  not  in  a 
straight  line,  these  angles  are  not  supplements  of  each  other. 

If  two  adjacent  angles  are  not  supplements  of  each  other, 
their  exterior  sid^s  are  not  in  the  same  straight  line. 


20 


PLANE   GEOMETRY.  —  BOOK   I. 


Peoposition  IV.     Theorem. 

95.  If  one  straight  line  intersects  another  straight 
line,  the  vertical  angles  are  equal.  , 


§90 

§90 


Let  line  OP  cut  AB  at  C. 

To  prove  Z  OCB  =  Z  AOF. 

Proof.  Z  OCA  +  Z  OCB  =  2  rt.  A, 

{being  sup.-adj.  A). 

Z0CA-\-ZA0P=2n  A, 

{being  sup.-adj.  A).  » 

.\Z  OCA  +  Z  OCB  =  Z  OCA  r  Z  ACF.        Ax.  1 
Take  away  from  each  of  these  equals  the  common  Z  OCA. 
Then  Z  OCB  =  Z  ACP.  Ax.  3 

In  like  manner  we  may  prove 

ZACO  =  ZFCB.  CIE.D. 

96.  Cor.  J^  one  of  the  four  angles  formed  hy  the  intersection 
of  two  straight  lines  is  a  right  angh^  the  other  three  angles  are 
right  angles. 


THE   STRAIGHT   LINE.  21 


Proposition  V.     Theorem. 


97.  From  a  -point  without  a  straight  line  one  per- 
pendicular, and  only  one,  can  he  drawn  to  this  line. 


Let  P  be  the  point  and  AB  t^e  line. 

To  prove  that  one  perpendicular,  and  only  one,  can  he  drawn 
from  P  to  AB. 

Proof.  Turn  the  part  of  the  plane  above  J. 7i  about  AB  as 
an  axis  until  it  falls  upon  the  part  below  AB,  and  denote  by 
/*'  the  position  that  P  takes. 

Turn  the  revolved  plane  about  AB  to  its  original  position, 
and  draw  the  straight  line  PP\  cutting  AB  at  C. 

Take  any  other  point  Dm  AB,  and  draw  PD  and  P'B. 

Since  PCP^  is  a  straight  line,  PDP  is  not  a  straight  line. 
{Between  two  points  only  one  straight  line  can  he  drawn.) 
.'.  A  PCP'iQ  a  St.  Z,  and  Z  PDP  is  not  a  st.  Z, 
Turn  the  figure  PCD  about  AB  until  P  falls  upon  P. 
Then  CP  will  coincide  with  CP,  and  DP  with  DP. 
,\^PCD  =  Z.PCD,2.ndiZ.PDC=Z.PDa  §59 

.-.  Z  PCD,  the  half  of  st.  Z  PCP,  is  a  rt.  Z  ;  and  Z  PDG, 
the  half  of  Z  PDP,  is  not  a  rt.  Z. 
.-.  PC  is  ±  to  ^^,  and  PD  is  not  J.  to  AB.  §  47 

.*.  one  X,  and  only  one,  can  be  drawn  from  Pto  AB. 


22  PLANE    GEOMETRY.  —  BOOK    I. 

Parallel  Lines. 

98.  Def.  Parallel  lines  are  lines  whicli  lie  in  the  same 
plane  and  do  not  meet  however  far  they  are  prolonged  in  both 
directions. 

99.  Parallel  lines  are  said  to  lie  in  the  same  direction  when 
they  are  on  the  same  side  of  the  straight  line  joining  their  ori- 
gins, and  in  opposite  directions  when  they  are  on  opposite  sides 
of  the  straight  line  joining  their  origins. 

Proposition  VI. 

100.  Two  strai0it  lines  in  the  same  plane  perpen- 
dicular  to  the  sume  straight  line  are  parallel. 


-B 


Let  AB  and  CD  be  perpendicular  to  AC. 

To  prove  AB  and  CD  parallel. 

Proof,   li  AB  and  CI)  are  not  parallel,  they  will  meet  if 

sufficiently  prolonged,  and  we  shall  have  two  perpendicular 

lines  from  their  point  of  meeting  to  the  same  straight  line ; 

but  this  is  impossible.  §  97 

[From  a  given  point  without  a  straight  line,  one  perpendicular,  and  only 
one,  can  be  drawn  to  the  straight  line.) 

.'.  AB  and  CD  are  parallel.  q.e.  d 

Remark.  Here  the  supposition  that  AB  and  CD  are  not  parallel  leads 
to  the  conclusion  that  two  perpendiculars  can  be  drawn  from  a  given 
point  to  a  straight  line.  The  conclusion  is  false,  therefore  the  supposi- 
tion iR  false;  but  if  it  is  false  that  AB  and  CD  are  not  parallel,  it  is  true 
that  they  are  parallel.  This  method  of  proof  is  called  the  indirect 
method. 

101.  Ax.  Through  a  given  point,  one  straight  line,  and  only 
one,  car  he  drawn  parallel  to  a  given  straight  line. 


PARALLEL   LINES. 


23 


Proposition  VII.    Theorem. 

102.  If  a  straight  line  is  perpendicular  to  one  of 
two  parallel  lines,  it  is  perpendicular  to  the  other. 


O 


M- 


C 


■N 


Let  AB  and  EF  be  two  parallel  lines,  and  let  HK  be 
perpendicular  to  AB. 

To  prove  IIK 1.  EF. 

Proof.    Suppose  JfiV' drawn  through  CI.  to  HK, 
Then  JfiVisllto^^,  §100 

{two  lines  in  the  same  plane  X  to  a  given  line  are  parallel). 

But  EFiQ  II  to  AB.  Hyp. 

.\  ^i^ coincides  with  J/iV,  §  101 

{through  the  same  point  only  one  line  can  he  drawn  ^  to  a  given  line). 
r.EFis±to  HK, 
that  is,  HK  is  ±  to  EF.  o.  e.  a 

103.  If  two  straight  lines  AB 
and  CD  are  cut  by  a  third  line 
EF^  called  a  transversal,  the 
eight  angles  formed  are  named 
as  follows : 

The  angles  a,  c?,/,  g  are  called 
intoior;  b,  c,  e,  h  are  called  ex- 
teiior  angles. 

The  angles  d  and  /,  or  a  and  g,  are  called  alU-int.  angles. 

The  angles  h  and  A,  or  c  and  e,  are  called  alt.-ext.  angles. 

The  angles/ and  h,  c  and  g,  a  and  e,  or  d  and  h,  are  called 
extAnt.  angles. 


24  PLANE   GEOMETRY.  —  BOOK   I. 


Proposition  VIII.     Theorem. 

104.  If  two  -parallel  straight  lines  are  cat  by  a  third 
straight  line,  the  alternate-interior  angles  are  equal. 


E- 


O. ^ ^-- H 

Let  EF  and  GH  be  two  parallel  straight  lines  cut  by 
the  line  BG, 

To  prove  Z  B  =  Z  C. 

Proof.  Through  0,  the  middle  point  of  £0,  suppose  AD 
drawn  J.  to  GIT. 

Then  AD  is  likewise  X  to  UF,  §  102 

(a  straight  line  ±  to  one  of  two  lis  is  ±  to  the  other), 

that  is,  CD  and  DA  are  both  JL  to  AD. 

Apply  figure  COD  to  figure  BOA,  so  that  OD  shall  fall 
on  OA. 

Then  0(7willfallon  OB, 

{since  Z.  COD  =•  Z.  BOA,  being  vertical  ^) ; 

and  the  point  C  will  fall  upon  -5, 

{since  00=  OB  by  construction). 

Then      the  JL  CD  will  coincide  with  the  ±  BA,  §  97 

{from  a  point  without  a  straight  line  only  one  ±  to  that  line  can  be  drawn). 

.*.  Z  OCD  coincides  with  Z  DBA,  and  is  equal  to  it.  §59 

^  Q.E.D. 

7TT  .  IP 

Ex.  1.  Find  the  value  of  an  angle  if  it  is  double  its  complement ;  if 
It  is  one-fourth  of  its  complement.        \.  -,a^ 

Ex.  2.  Find  the  value  of  an  angle  if  itS«  double  its  supplement ;  if  it 
18  one-third  of  its  supplement. 


PARALLEL  LIKES.  26 


Proposition  IX.    Theorem. 

105.  Conversely  :  When  two  straight  lines  are  cut 
by  a  third  straight  line,  if  the  alternate  interior  aw 
gles  are  equal,  the  two  straight  lines  are  parallel. 


N 


Let  EF  cut  the  straight  lines  AB  and  CD  in  the  points 
U  and  A',  and  let  the  Z.AUK  =  ^HKD. 

To  prove  A  B  Ho  CD. 

Proof.     Suppose  ifi\r drawn  through  H\\  to  CD;         §  101 

then  AMHK=Z.HKD,  §104 

{being  alt.-int  A  ofW  lines). 

But  ^AHK=Z.HKD;  Hyp. 

.-.  Z.  MHK=  Z  AHK,  .  Ax.  1 

/.  the  lines  JIfiVand  AB  coincide. 
But  J/iVis  II  to  CD,  Cons. 

•'.  AB,  which  coincides  with  MIT.  is  II  to  CD. 


Ex.  3.   IIow  many  degrees  in  the  angle  formed  by  the  hands  of  a 
clock  at  2  o'clock  ?  3  o'clock  ?  4  o'clock  ?   6  o'clock  ? 


26  PLANE    GEOMETRY.  —  BOOK    I, 


Proposition  X.     Theorem. 

106.  If  two  parallel  lines  are  cut  by  a  third  straight 
y^e,  the  exterior-interior  angles  are  equal, 

E 

/ 


Let  AB  and  CD  be   two  parallel  lines   cut  by  the 
straight  line  EF^  in  the  points  H  and  K, 

To  'prove  Z  EHB  =  A  HKD. 

Proof.  Z  EHB  =  Z  Alfl^,  §  95 

(being  vertical  A). 

But  Z  AirK=  Z  irXB,  §  104 

(being  alt.-int.  AqfW  lines). 

,\  Z  EITB  =  Z  HKD.  Ax.  1 

In  like  manner  we  raay  prove 

Z  EHA  =  Z  HKC. 

Q.  E.  D. 

107.    Cor.    The  alternate-exterior  angles   EHB  and  CKF, 
and  also  AHE  and  DKF,  are  equal. 


Ex.  4.  If  an  angle  is  bisected,  and  if  a  line  is  drawn  through  the 
vertex  perpendicular  to  the  bisector,  this  line  forms  equal  angles  with 
the  sides  of  the  given  angle. 

Ex.  5.  If  the  bisectors  of  two  adjacent  angles  are  perpendicular  to 
each  other,  the  adjacent  angles  are  supplementary. 


PAEALLEL   LINES.  27 


Proposition  XI.     Theorem. 

108.  Conversely  :  When  two  straight  lines  are  cut 
hy  a  third  straight  line,  if  the  exterior-interior  an- 
gles are  equal,  these  two  straight  lines  are  parallel. 


Let  EF  cut  the  straight  lines  AB  and  CD  in  the  points 
H  and  K,  and  let  the  Z  EI/B  =  Z  HKD. 

To  prove  ^5  II  to  CD. 

Proof.     Suppose  J/iV  drawn  through  H II  to  CD.         §  101 

Then  /.  EHN=  Z  HKD,  §  106 

{being  ext.-int.  AofW  lines). 

But  Z  UJIB  =  Z  JIKD.  Hyp. 

,\Z.EIIB=^Z.EHN.  Ax.  1 

/.the  lines  JfiV and  AB  coincide. 
But  MN\^  II  to  CD.  Cons. 

.'.  AB,  which  coincides  with  MN,  is  II  to  CD. 

Q.  E.  D. 


Ex.  6.  The  bisector  of  one  of  two  vertical  angles  bisects  the  other. 
Ex.  7.   The  bisectors  of  the  two  pairs  of  vertical  angles  formed  by 
two  intersecting  lines  are  perpendicular  to  each  other. 


28  PLANE  GEOMETRY.  —  BOOK   I. 


Proposition  XII.    Theorem. 

109.  If  two  -parallel  lines  are  cut  hy  a  third  straight 
line,  the  sum  of  the  two  interior  angles  on  the  same 
side  of  the  transversal  is  equal  to  two  right  angles. 


Let  AB   and  CD  be   two  parallel  lines  cut  by  the 
straight  line  EF  in  the  points  H  and  K, 

To  prove         Z  BHK-^  Z  HKD  =  2  rt.  A. 

Proof.  Z  U:ff£  +  Z  BirK=  2  rt.  A,  §  iK) 

{being  sup.-adj.  A). 

But  /.EB:B^Z.IIKD,       /  §106 

{being  ext-int.  A  of  II  lines). 

Substitute  Z  HKD  for  Z  EHB  in  the  first  equality  ; 

then  Z  BHK-\-  Z  HKD  =  2  rt.  A. 


Ex.  8.   If  the  angle  AHE  is  an  angle  of  135°,  find  the  number  of 
degrees  in  each  of  the  other  angles  formed  at  the  points  //and  K, 

Ex.  9.    Find  the  angle  between  the  bisectors  of  adjacent  com})lemen- 
tary  angles. 


PARALLEL   LINES.  ^  29 


Proposition  XIII.     Theorem. 

110.  Conversely  :  WJien  two  straight  lines  are  cut 
by  a  third  straight  line,  if  tlve  two  interior  angles  on 
the  same  side  of  the  transversal  are  together  equal  to 
two  right  angles,   then  the  two  straight  lines  are 

parallel. 


Let  EF  cut  the  straight  lines  AB  and  CD  in  the  points 
n  and  K,  and  let  the  Z  BHK  +  Z  JIKD  equal  two  right 
angles. 

To  prove  AB  II  to  CD. 

Proof.     Suppose  MN  drawn  through  IT  II  to  CD. 

Then  Z  NUK-h  Z  ITKD  =  2  rt.  A,  §  109 

{being  two  interior  A  of  lis  on  tJie  same  side  of  the  transversal). 

But  Z  BJIX-\-ZirXI)  =  2  rt.  A.  Hyp. 

.'.Z.NHK^Z.HKD  =  /.BHK-^AHKn.    Ax.  1 

Take  away  from  each  of  these  equals  the  common  Z  HKD ; 

then  Z  NHK=  Z  BHK.  Ax.  3 

.'.  the  lines  AB  and  JI/TV coincide. 

But  JfiVis  II  to  CD.  Cons. 

.'.  AB,  which  coincides  with  MN,  is  II  to  CD. 

aE.  D. 


30 


PLANE  GEOMETRY. 


BOOK   I. 


Proposition  XIV.     Theorem. 

111.  Two  straight  lines  which  are  parallel  to  a  third 
straight  line  are  parallel  to  each  other. 


E- 


K 


Let  AB  and  CD  be  parallel  to  EF, 

To  prove  AB  W  to  CD. 

Proof.  Suppose  ^^  drawn  JL  to  ^i^.  §97 

Since  CD  and  ^i^are  II,  UK  h  JL  to  CD,        §  102 

{if  a  straight  line  is  ±  to  one  of  two  Us,  it  is  ±  to  the  other  also). 

Since  AD  and  ^i^are  II,  JIKis  also  X  to  AD.    §  102 

.',ZirOD  =  ZIfFD, 
(each  being  a  rt  Z). 
.-.  AD  is  II  to  CD,  §  108 

{when  two  straight  lines  are  cut  by  a  third  straight  line,  if  the  ext.-int.  A 
are  equal,  the  two  lines  are  parallel). 

a  E.  D. 

Ex.  10.  It  has  been  shown  that  if  two  parallels  are  cut  by  a  trans- 
versal, the  alternate-interior  angles  are  equal,  the  exterior-interior  angles 
are  equal,  the  two  interior  angles  on  the  same  side  of  the  transversal  are 
supplementary.  State  the  opposite  theorems.  State  the  converse  theo- 
rems. 


PARALLEL    LINES.  31 


Proposition  XV.    Theorem.  /  i- 


112.  Two  angles  whose  sides  are  parallel  each   to 
ea^h,  are  either  equal  or  supplementary. 


Let  AB  be  parallel  to  EF,  and  BC  to  MN. 

To  p'ove  Z.  ABC  equal  to  Z  EHN,  and  to  Z.  MEF,  and 
suppletnentary  to  Z.  EHM  and  to  Z  NHF. 

Proof.   Prolong  (if  necessary)  5(^and  i^^  until  they  inter- 
sect at  D.  •  §  81  (2) 
Then                          ZB^ZEDC,  §106 

and  Z  DB:N=  Z  EDO, 

{being  ext.-int.  AqfW  lines), 

,\ZE  =  ZI)IIN;  Ax.  1 

and  ZB  =  Z  MHF  (the  vert.  Z  of  DHN). 

Now  Z  DHN'i^  the  supplement  of  Z  EE:3f  and  Z  NHF. 

.-.  Z  B,  which  is  equal  to  Z  DHN, 

is  the  supplement  of  Z  EHIfand  of  Z  NHF. 

Q.E.  D. 

Remark.  The  angles  are  equal  when  both  pairs  of  parallel  sides 
extend  in  the  same  direction,  or  in  opposite  directions,  from  their  ver- 
tices ;  the  angles  are  supplementary  when  two  of  the  parallel  sides  extend 
in  the  same  direction,  and  the  other  two  in  opposite  directions,  from  theii 
vertices. 


32 


PLANE   GEOMETRY.  —  BOOK   I. 


Proposition  XVI.    Theorem. 

113.  Two  angles  whose  sides  are  perpendicular  each 
to  each,  are  either  equal  or  supplementary. 

G 

K 


Let  AB  be  perpendicular  to  FD,  and  AG  to  GI. 

To  prove  Z  BAQ  equal  to  Z  DFG,  and  supplementary  to 
Z.DFL 

Proof.   Suppose  J.^ drawn  ±  to  AB^  and  AH  1.  to  AQ. 

Then  A^is  11  toi^D,  and^^to76^,  §100 

{^wo  lines  J-  to  the  same  line  are  parallel). 

.\ZI)FG  =  ZKAir,  §112 

{two  angles  are  equal  whose  sides  are  II  and  extend  in  the  same  direction 
from  their  vertices). 

The  Z  BAKis  a  right  angle  by  construction. 

.*.  Z  BAHis  the  complement  of  Z  KAH. 
The  Z  C^^is  a  right  angle  by  construction. 

.*.  Z  BAHis  the  complement  of  Z  BAC. 

.\ZBAC=ZKAir,  §87 

(complements  of  equal  angles  are  equal). 

.\ZDFO-^ZBAC.  Ax.  1 

/.  Z  DFI,  the  supplement  of  Z  DFG,  is  also  the  supplement 

ofZ^^C.  aE.D. 

Remark.   The  angles  are  equal  if  both  are  acute  or  both  obtuse ;  they 
are  supplementary  if  one  is  acute  and  the  other  obtuse. 


PERPENDICULAR   AND   OBLIQUE   LINES. 


33 


Perpendicular  and  Oblique  Lines. 

Proposition  XVII.    Theorem.  y 

114.   The  perpendicular  is  the  shortest  line  that  can 
he  drawn  from  a  point  to  a  straight  line. 


Let  AB  he  the  given  straight  line,  P  the  given  point, 
PC  the  perpendicular,  and  PD  any  other  line  drawn 
from  P  to  AB. 

To  prove  PC  <  PD. 

Proof.  Produce  PC  to  P\  making  CP^=  PC;  and  draw  DP\ 
On  AB  as  an  axis,  fold  over  CPD  until  it  comes  into  the 
plane  of  CP'D. 

The  line  CP  will  take  the  direction  of  CP\ 
{since  Z  PCD  =-ZP^CD,  each  being  a  rt.  Z). 
The  point  P  will  fall  upon  the  point  P\ 
{since  PC=  FChy  cons.). 
,'.  line  PD  =  line  P'A 
^'.  PZ>4-P'i>  =  2Pi), 
and  PC  -\-CP'  =2  PC  Cons. 

But  PC  +  CP'  <PB+  DP\ 

{a  straight  line  is  the  shortest  distance  between  two  points). 

.-.  2PC<  2PD.  or  PC<  PP.  Q.E.a 


34  PLANE   GEOMETRY.  —  BOOK   I. 

115.  Scholium.  The  distance  of  a  point  from  a  line  is  under- 
stood to  mean  the  length  of  the  perpendicular  from  the  point 
to  the  line. 

Proposition  XVIII.     Theorem. 

116.  Two  oblique  lines  drawn  from  a  point  in  a 
perpendicular  to  a  given  line,  cutting  off  equxil  dis- 
tances from  the  foot  of  the  perpendicular,  are  equal. 


Let  FC  "be  the  perpendicular,  and  CA  and  CO  two 
oblique  lines  cutting  off  equal  distances  from  F. 

To  prove  CA  =  CO. 

•Proof.    Fold  over  CFA,  on  CF&s  an  axis,  until  it  comes  into 
the  plane  of  CFO. 

FA  will  take  the  direction  of  FO, 
{since  Z  CFA  =  Z  CFO,  each  being  a  rt.  Z  by  hyp.). 

Point  A  will  fall  upon  point  0, 
{since  FA  =  FO  by  hyp.). 
.-.  line  CA  =  line  CO, 
(their  extremities  being  the  same  points).  q.  e.  d. 

117.  Cor.  Two  oblique  lines  drawn  from  a  point  in  a  per- 
pendicular to  a  given  line,  cutting  off  equal  distances  from  the 
foot  of  the  perpendicular,  make  equal  angles  with  the  given  line. 
and  also  with  the  perpendicular. 


PERPENDICULAR   AND   OBLIQUE   LINES.  35 


Proposition  XIX.    Theorem. 

118.  The  sum  of  two  lines  drawn  from  a  point  to 
the  extremities  of  a  straight  line  is  greater  than  the 
sum  of  two  other  lines  similarly  drawn,  hut  included 
hy  them. 


Let  CA  and  OB  be  two  lines  drawn  from  the  point  C 
to  the  extremities  of  the  straight  line  AB.  Let  OA  and 
OB  be  two  lines  similarly  drawn,  but  included  by  CA 
and  OB. 

To  prcyve  CA-{-CB>OA  +  OB. 

Proof.      Produce  AO  to  meet  the  line  CB  at  E. 

Then  AC'\-OE>  0A-\- OE, 

(a  straight  line  m  the  shortest  distance  between  two  "pointfl^ 

and  BE-\-OE>  BO. 

Add  these  inequalities,  and  we  have 

CA  +  CE-\-  BE-\-OE>OA-\-  0E-\-  OB, 

Substitute  for  CE-\-  BE  its  equal  CB, 

and  take  away  OE  from  each  side  of  the  inequality. 

We  have  CA  +  CB>OA-\-  OB.  Ax.  5    q.  e.  d. 


36  PLANE   GEOMETRY.  —  BOOK   I. 

Proposition  XX.     Theorem. 

119.  Of  two  oblique  lines  drawn  from  the  same 
point  in  a  perpendicular,  cutting  off  unequal  dis- 
tances from  the  foot  of  the  perpendicular,  the  more 
remote  is  the  greater. 


\ ,  D 

Let  OC  be  perpendicular  to  AB,  OG  and  OE  two  oblique 
lines  to  AB,  and   GE  greater  than  CG. 

To  prove  OE  >  00. 

Proof.        Take  OF  equal  to  CG,  and  draw  OF. 

Then  OF=OG,  §116 

{two  oblique  lines  drawn  from  a  point  in  a  ±,  cutting  off  equal  distances 
from  the  foot  of  the  ±,  are  equal). 

Prolong  OC  to  D,  making  CD  =  OC. 

Draw  ED  and  ED. 

Since  AB  is  J_  to  OD  at  its  middle  point, 

EO  =  ED,  and  EO  =  ED,  §  116 

But  OE-\-ED>  0E-{-  ED,  §  118 

{the  sum  of  two  oblique  lines  drawn  from  a  point  to  the  extremities  of  a 
straight  line  is  greater  than  the  sum  of  two  other  lines  similarly  drawn, 
but  included  by  them). 

.-.  20E>  20E,  or  OE >  OE 
But  0E=  OG.     Hence  OE  >  OG.  q.  e.  d. 

120.  Cor.  Only  two  equal  straight  lines  can  be  drawn  from 
a  point  to  a  straight  line ;  and  of  two  unequal  lines,  the  greater 
cuts  off  the  greater  distance  from  the  foot  of  the  perpendicular. 


PERPENDICULAR   AND   OBLIQUE    LINES.  37 


Proposition  XXL    Theorem. 

121.  Two  equal  oblique  lines,  drawn  from  the  same 
point  in  a  perpendicular,  cut  off  equal  distances  from 
the  foot  of  the  perpendicular. 


Let  CF  be  the  perpendicular,  and  CE  and  CK  be  two 
equal  oblique  lines  drawn  from  the  point  C  to  AB. 

To  prove  FE=  FK. 

Proof,  Fold  over  CFA  on  (TFas  an  axis,  until  it  comes  into 
the  plane  of  CFB. 

The  line  FE  will  take  the  direction  FK, 
{since  Z  CFE  =  Z  CFK,  each  being  a  rt.  Z  by  hyp.). 

y        Then  the  point  E  must  fall  upon  the  point  K, 

zxi^FE^FK. 

Otherwise  one  of  these  oblique  lines  must  be  more  remote 
from  the  perpendicular,  and  therefore  greater  than  the  other ; 
which  is  contrary  to  the  hypothesis  that  they  are  equal.  §  119 

a  E.  D. 


Ex.  11.  Show  that  the  bisectors  of  two  supplementary  adjacent 
angles  are  perpendicular  to  each  other. 

Ex.  12.  Show  that  the  bisectors  of  two  vertical  angles  form  one 
straight  line. 

Ex.  13.  Find  the  complement  of  an  angle  containing  26°  52''  37''''. 
Find  the  supplement  of  the  same  angle. 


38 


PLANE   GEOMETRY. — BOOK   I. 


Proposition  XXII.     Theorem. 

122.  Every  point  in  the  perpendicular,  erected  at 
the  jniddle  of  a  given  straight  line,  is  equidistant 
from  the  extremities  of  the  line,  and  every  point  not 
ijv  the  perpendicular  is  unequally  distant  from  the 
extremities  of  the  line. 


Let  PR  be  a  perpendicular  erected  at  the  middle  oi 
the  straight  line  AB,  0  any  point  in  PR,  and  C  any 
point  without  PR. 

Draw  OA  and  OB,  CA  and  CB. 
To  prone  OA  and  OB  equal,  CA  and  CB  unequal. 
Proof.  PA  =  PB.  Hyp. 

,'.OA  =  OB,  §116 

{two  oblique  lines  drawn  from  the  same  point  in  a  ±,  cutting  off  equal  dis- 
tances from  the  foot  of  the  ±,  are  equal). 

Since  C  is  without  the  perpendicular,  one  of  the  lines,  CA 

or  CB,  will  cut  the  perpendicular. 

Let  CA  cut  the  _L  at  I),  and  draw  PB. 

Then  PB  =  PA, 

{two  oblique  lines  drawn  from  the  same  point  in  a  ±,  cutting  off  equal  dis- 
tances from  the  foot  of  the  ±,  are  equal). 

But  CB<CP  +  PB, 

{a  straight  line  is  the  shortest  distance  between  two  points). 
Substitute  in  this  inequality  PA  for  PB,  and  we  have 

CB<CP-{-PA. 
That  is,  CB  <  CA,  0.6.^ 


PERPENDICULAR   AND   OBLIQUE    LINES.  39 

123.  Since  two  points  determine  the  position  of  a  straight 
line,  two  pninfji  eqiddn^tani  frnm-ih^£xtrp,raitip.R  of  a  line  deter- 
mine the  perpendicular  at  the  middle  of  that  line. 

The  Locus  of  a  Point. 

124.  If  it  is  required  to  find  a  point  which  shall  fulfil  a 
single  geometric  condition,  the  point  will  have  an  unlimited 
number  of  positions,  but  will  be  confined  to  a  particular  line, 
or  group  of  lines. 

Thus,  if  it  is  required  to  find  a  point  equidistant  from  the 
extremities  of  a  given  straight  line,  it  is  obvious  from  the  last 
proposition  that  any  point  in  the  perpendicular  to  the  given 
line  at  its  middle  point  does  fulfil  the  condition,  and  that  no 
other  point  does ;  that  is,  the  required  point  is  confined  to  this 
perpendicular.  Again,  if  it  is  required  to  find  a  point  at  a 
given  distance  from  a  fixed  straight  line  of  indefinite  length,  it 
is  evident  that  the  point  must  lie  in  one  of  two  straight  lines, 
so  drawn  as  to  be  everywhere  at  the  given  distance  from  the 
fixed  line,  one  on  one  side  of  the  fixed  line,  and  the  other  on 
the  other  side. 

The  lociis  of  a  point  under  a  given  condition  is  the  line, 
or  group  of  lines,  which  contains  all  the  points  that  fulfil  the 
given  condition,  and  no  other  points. 

125.  Scholium.  In  order  to  prove  completely  that  a  certain 
line  is  the  locus  of  a  point  under  a  given  condition,  it  is  neces- 
sary to  prove  that  every  point  iyi  the  line  satisfies  the  given 
condition;  and  secondly,  that  every  point  which  satisfies  the 
given  condition  lies  in  the  line  (the  converse  proposition),  or 
that  every  point  not  in  the  line  does  not  satisfy  the  given  condi- 
tion (the  opposite  proposition). 

126.  Cor.  The  locus  of  a  point  equidistant  from  the  extrem- 
ities of  a  straight  line  is  the  perpendicular  bisector  of  that  line. 

§§  122,  123 


40 


PLANE   GEOMETEY.  —  BOOK   I. 


Triangles. 

\y    127.   A  triangle  is  a  portion  of  a  plane  bounded  by  three 
straight  lines;  as,  ABC. 

The  bounding  lines  are  called  the 
sides  of  the  triangle,  and  their  sum  is 
called  its  pei'imeter ;  the  angles  formed 
by  the  sides  are  called  the  angles  of  the 
triangle,  and  the  vertices  of  these  an- 
gles, the  vertices  of  the  triangle. 

128.  An  exterior  angle  of  a  triangle 
is  an  angle  formed  between  a  side  and 
the  prolongation  of  another  side ;  as, 
ACD.  The  interior  angle  ACB  is 
adjacent  to  the  exterior  angle ;  the 
other  two  interior  angles,  A  and  B,  are  called  opposite- 
interior  angles. 


Scalene. 


Isosceles. 


Equilateral. 


129.  A  triangle  is  called,  with  reference  to  its  sides,  a 
scalene  triangle  when  no  two  of  its  sides  are  equal;  an  isos- 
celes triangle,  when  two  of  its  sides  are  equal ;  an  equilateral 
triangle,  when  its  three  sides  are  equal. 


Right.      •  Obtus.e. 

130.   A  triangle  is  called,  with  referen 
triangle,  when  one  of  its  angles  is  a  right  angle ;  an  obt/use 


Acute.  Equiangular. 

ce  to  its  angles,  a  right 


TRIANGLES.  i  41 

triangle^  when  one  of  its  angles  is  an  obtuse  angle ;  an  acute 
triangle,  when  all  three  of  its  angles  are  ac ite  angles;  an 
equiangular  triangle,  when  its  three  angles  are  equal. 

13L  In  a  right  triangle,  the  side  opposite  the  right  angle  is 
called  the  hypotenuse,  and  the  other  two  sides  the  legs,  of  the 
triangle. 

132.  The  side  on  which  a  triangle  is  supposed  to  stand  is 
called  the  base  of  the  triangle.  In  the  isosceles  triangle,  the 
equal  sides  are  called  the  legs,  and  the  other  side,  the  base  ;  in 
other  triangles,  any  one  of  the  sides  may  be  taken  as  the  base. 

133.  The  angle  opposite  the  base  of  a  triangle  is  called  the 
vertical  angle,  and  its  vertex  the  vertex  of  the  triangle. 

134.  The  altitude  of  a  triangle  is  the  perpendicular  distance 
from  the  vertex  to  the  base,  or  to  the  base  produced ;  as,  AD. 

135.  The  three  perpendiculars  from  the  vertices  of  a  tri- 
angle to  the  opposite  sides  (produced  if  necessary)  are  called 
the  altitudes;  the  three  bisectors  of  the  angles  are  called  the 
bisectors;  and  the  three  lines  from  the  vertices  to  the  middle 
points  of  the  opposite  sides  are  called  the  medians  of  the 
triangle. 

136.  If  two  triangles  have  the  angles  of  the  one  equal  respec- 
tively to  the  angles  of  the  other,  the  equal  angles  are  called 
homologous  angles,  and  the  sides  opposite  the  equal  angles  are 
called  homologous  sides. 

In  general,  .points,  lines,  and  angles,  f,imilarly  situated  in 
equal  or  similar  figures,  are  called  homologous. 

137.  Theorem.  The  sum  of  two  sides  of  a  triangle  is  greater 
than  the  third  side,  and  their  difference  is  less  thafi  the  third 
side. 

In  the  A  ABC {Y\g.  1),  AB-\-BO>AO,  for  a  straight  line 
is  the  shortest  distance  between  two  points ;  and  by  taking 
away  ^Cfrom  both  sides,  AB>AC-BO,oi  AC-BC<AB. 


42  PLANE   GEOMETRY.  —  BOOK   I. 


Proposition  XXIII.    Theorem. 

138.  The  sujvb  of  the  three  angles  of  a  triangle  is 
equal  to  two  right  angles. 


A  C " F 

Let  ABC  be  a  triangle. 

To  prove       Z  B -\- Z  BCA  -\-ZA^2xi.  A. 

Proof.    Suppose  (7^  drawn  II  to  AB,  and  prolong  AC io  F. 

Then       Z  ECF+  Z  ECB  +  Z  BOA  =  2  rt.  A,  §  92 

{the  sum  of  all  the  A  about  a  point  on  the  same  side  of  a  straight  line 

=  2  rt.  A). 

But  ZA  =  Z  EOF,  §  106 

(being  ext.-int.  AofW  lines). 

8indLZB  =  ZBCE,  §104 

(being  alt.-int.  A  of  II  lines). 

Substitute  for  Z  EOF  SLud  Z  BCE  the  equal  A  A  and  B. 

Then  Z  A+Z  B +  Z  BCA  =  2  rt  A. 

aE.  D. 

139.  Cor.  1.  If  the  sum  of  two  angles  of  a  triangle  is  sub- 
tracted from  two  right  angles,  the  remainder  is  equal  to  the 
third  angle. 

140.  Cor.  2.  If  two  triangles  have  two  angles  of  the  one 
equal  to  two  angles  of  the  other,  the  third  angles  are  equal. 

141.  Cor.  3.  If  two  right  triangles  have  an  acute  angle  of 
the  one  equal  to  an  acute  angle  of  the  other,  the  other  acute 
anglA  are  equal.  , 


TRIANGLES. 


43 


142.  Cor.  4.  In  a  tiiangle  then-e  can  he  bid  one  right  angU^ 
or  one  obtuse  angle. 

143.  Cor.  b.  In  a  right  triangle  the  two  acute  angles  are 
complements  of  each  other. 

144.  Cor.  6.  In  an  equiangular  triangle,  each  angle  is  one- 
third  of  two  right  angles,  or  two-thirds  of  one  right  angle. 

Proposition  XXIV.     Theorem. 

145.  The  exterior  angle  of  a  triangle  is  equal  to  the 
sum  of  the  two  opposite  interior  angles. 


Let  BCH  be  an.  exterior  angle  of  the  triangle  ABC. 
To  prove  Z.  BOH^  A  A-\- /.  B. 

Proof.  Z.BCH^AACB  =  2x\..  A 

{being  sup.-adj.  A). 

ZA  +  AB  +  Z  ACB  =  2  rt.  A,  §  138 

{the  sum  of  the  three  A  of  a  A  =  2  rt.  A). 

,\  Z  BCH-\-  Z  ACB  =  Z  A  +  Z  B  -\-  Z  ACB.     Ax.  1 

Take  away  from  each  of  these  equals  the  common  Z  A  CB ; 

then  ZBGH=.^^A-\-ZB.  Ax.  3 

Q.E.O. 

146.   Cor.    The  exterior  angle  of  a  triangle  is  greater  than 
either  of  the  opposite  interior  angles. 


44  PLANE   GEOMETRY.  —  BOOK   I. 


Proposition  XXV.     Theorem. 

147.  Two  triangles  are  equal  if  a  side  and  two  ad- 
jacent anglers  of  the  one  are  equal  respectively  to  a 
side  and  two  adjacent  angles  of  the  other. 


In  the  triangles  ABC  and  DEF,  let  AB  =  DE,  ZA=:ZD, 
/.B^AE. 

To  'prove  A  ABC  =  A  DEF. 

Proof.  Apply  the  A  ABC  to  the  A  DEF  so  that  AB  shall 
coincide  with  BE. 

A  C  will  take  the  direction  of  DF, 
{for  Z  A  =  Z  D,  by  hyp.) ; 

the  extremity  C  of  J.  C  will  fall  upon  BF  or  BE  produced. 

BC  will  take  the  direction  of  EF, 
(for  ZB  =  ZB,by  hyp.) ; 

the  extremity 'C  of  BC  will  fall  upon  EF  or  ^i^  produced. 

.'.the  point  C,  falling  upon  both  the  lines  BF  smd  EF, 
must  fall  upon  the  point  common  to  the  two  lines,  namely,  F. 

/.the  two  A  coincide,  and  are  equal.  q. e. d. 

148.  Cor.  1.  Two  riyht  triangles  are  equalif  the  hypotenuse 
and  an  acute  angle  of  the  one  are  equal  respectively  to  the  hypote- 
nuse and  an  acute  angle  of  the  other. 

149.  Cor.  2.  Two  right  triangles  are  equal  if  a  side  and  an 
acute  angle  of  the  one  are  equal  respectively  to  a  side  and 
homologous  acute  angle  of  the  other. 


TRIANGLES. 


45 


Proposition  XXVI.     Theorem. 

150.  Two  triangles  are  equal  if  two  sides  and  the 
included  angle  of  tJie  one  are  equal  respectively  to 
two  sides  and  the  included  angle  of  the  other. 


D 


In  the  triangles  ABC  and  DEF,  let  AB  =  DE,  AC  =  DF, 
ZA  =  ZD. 


To  prove 


AABC^ADEF. 


Proof.   Apply  the  A  ABC  to  the  A  DEF  so  that  AB  shall 
coincide  with  DE. 

Then  ^Cwill  take  the  direction  of  DF, 

{forZA  =  ZD,byhyp.); 

the  point  C  will  fall  upon  the  point  F, 
(JorAC=DF,byhyp.).'' 

.',CB  =  FE, 

{their  extremities  being  the  same  points). 

.'.the  two  A  coincide,  and  are  equal. 

Q.E.D. 

151.   Cor.    Two  right  triangles  are  equal  if  their  legs  arfi 
equal,  each  to  each. 


46 


PLANE    GEOMETRY. 


BOOK    I. 


Proposition  XXVIL     Theorem. 

162,  If  two  triangles  have  two  sides  of  the  one  equal ^ 
respectively  to  two  sides  of  the  other,  hut  the  included 
angle  of  tl%e  first  greater  than  the  included  angle  of 
the  second,  then  the  third  side  of  the  first  will  he 
greater  than  the  third  side  of  the  second. 


B 


^<:^^ 


ak 


#: 


1 


In  the  triangles  ABC  and  ABE,  let  AB  =  AB,  BC=BE; 
but  /.ABC  greater  than  /.ABE. 

To  prove  A^  >  AE. 

Proof.    Place  the  A  so  that  AB  of  the  one  shall  coincide  with 
AB  of  the  other. 

Suppose  ^i^  drawn  so  as  to  bisect  4-  EBQ. 
Draw  EF, 

In  the  A  EB F  s^ndi  QBE 

EB  =  BC,  Hyp. 

BF=BF,  Iden. 

Z.EBE=Z.CBF.  Cons. 

.-.  the  A  EBE  and  CBEine  equal,  §  150 

{having  two  sides  and  the  included  Z  of  one  equal  respectively  to  two  sides 
and  the  included  Z  of  the  other). 

.'.  EE=  EC, 

{being  homologous  sides  of  equal  A). 

Now  •  AE+EE>  AE,  §137 

{the  sum  of  two  sides  of  a  A  is  greater  than  the  third  side). 
.'.  AEi-EO  AE; 

or,  AC>  AE.  <^E.tt 


TRIANGLES.  47 


Proposition  XXVIII.     Theorem. 

153.  Conversely.  If  two  sides  of  a  triangle  are  equal 
respectively  to  two  sides  of  another,  hut  the  third  side 
of  the  first  triangle  is  greater  than  the  third  side  of 
the  second,  then  the  angle  opposite  the  third  side  of 
the  first  triangle  is  greater  than  the  angle  opposite 
the  third  side  of  the  second. 

D 


B  C  K  F 

In  the  triangles  ABC  and  DBF,  let  AB  =  DE,  AC  =  DF, 
but  let  BG  be  greater  than  FF, 
To  prove  Z  ^4  greater  than  Z  D. 

Proof.    Now  Z.  A   \^  equal  to  A  D,  or  less  than  Z  Z),  or 
greater  than  Z.  D. 

But  Z  ^  is  not  equal  to  Z  D,  for  then  A  ABC  would  be 
equal  to  A  DBF,  §  150 

ijuiving  two  sides  and  the  included  Z.  of  the  owe  respectively  equal  to  two 
aides  and  the  included  Z  of  the  other\ 

and  ^C  would  be  equal  to  EF. 

And  Z  ^  is  not  less  than  Z  D,  for  then  BC  would  be  less 
than  EF.  §  152 

.-.  Z  ^  is  greater  than  Z  B. 


48 


PLANE   GEOMETRY. — BOOK    I. 


Peoposition  XXIX.     Theorem. 

154.  In  an  isosceles  triangle  the  angles  opposite  the 
equal  sides  are  equal. 

A 


B  T)  C 

Let  ABO  be  an  isosceles  triangle,  having  the  sides 
AB  and  AC  equal. 

To  prove  Z  B  =  Z  C. 

Proof.    Suppose  AD  drawn  so  as  to  bisect  the  Z  BAC. 

In  the  A  ABB  and  ABC, 

AB-^AC.  Hyp. 

AD  =  AD,  Iden. 

Z  BAB  =  Z  CAD.  Cons. 

.\  A  ABB  =  A  ABC,  §150 

{two  A  are  equal  if  two  sides  and  the  included  Z  of  the  one  are  equal 
respectively  to  two  sides  and  the  included  Z  of  the  other). 

»'»  Z  B  =  Z  C,  Q.  E.  D. 

155.  Cor.  An  equilateral  triangle  is  equiangular,  and  each 
angle  contains  60°.  

Ex.  14.  The  bisector  of  the  vertical  angle  of  an  isosceles  triangle 
bisects  the  base,  and  is  perpendicular  to  the  base. 

Ex.  15.  The  perpendicular  bisector  of  the  base  of  an  isosceles  triangle 
passes  through  the  vertex  and  bisects  the  angle  at  the  vertex. 


i^RTANOLES. 


49 


Proposition  XXX.    Theorem. 

156.  If  two  angles  of  a  triangle  are  equal,  the  sides 
opposite  the  equal  angles  are  equal,  and  the  triangle 
is  isosceles. 


In  the  triangle  ABC,  let  the  ZB  =  ZC. 

To  prove  AB  =  AC. 

Proof.  Suppose  AD  drawn  _L  to  BC, 

In  the  rt.  A  ADB  jxnd  ADC, 

AD  =  AD,  Iden. 

ZB^ZC  Hyp. 

.-.  rt.  A  ADB  =  rt.  A  ADC,  §  149 

{having  a  side  and  an  acute. Z.  of  the  one  equal  respectively  to  a  side  and 
an  homologous  acute  Z  of  the  other). 

.\AB  =  AC, 

{being  homologous  sides  of  equal  A). 


aE.D. 


157.   Cor.    An  equiangular  triangle  is  also  equilateral. 


Ex.  16.  The  perpendicular  from  the  vertex  to  the  base  of  an  isosdfeles 
triangle  is  an  axis  of  symmetry.  ^^,„^ 


OO  PLANE    GEOMETRY..  —  BOOK    I. 

Proposition  XXXI.     Theorem. 

158.  If  two  sides  of  a  triangle  are  unequal,  the  an- 
gles opposite  are  unequal,  and  the  greater  angle  is 
opposite  the  greater  side. 


C  B 

In  the  triangle  AGB  let  AB  be  greater  than  AC, 

To  prove  Z  A  CB  greater  than  Z  B, 

Proof.  Take  AE  equal  to  AC. 

Draw  EC. 

ZAEC=ZACE,  §154 

{being  A  opposite  equal  sides). 

But  Z  ^^C  is  greater  than  Z^,  §146 

(an  exterior  Z  of  a  A  is  greater  than  either  opposite  interior  Z). 

and  Z  ACB  is  greater  than  Z  ACE.  Ax.  8 

Substitute  for  Z  ACE  its  equal  Z  A  EC, 

then  Z  ACB  is  greater  than  Z  A  EC. 

Much  more, then, is  the  Z  ACB  greater  than  Z  B. 

aE.D. 


Ex.  17.  If  the  angles  ABC  And  ACB,  at  the  base  of  an  isosceles  tri- 
angle, be  bisected  by  the  straight  lines  BD,  CD,  show  that  BBCmW 
be  an  isosceles  triangle. 


TRIANGLES.  61 


Proposition  XXXII.        Theorem.     « 

159.  Conversely  :  If  two  angles  of  a  triangle  are 
unequal,  the  sides  opposite  are  unequal,  and  the 
greater  side  is  opposite  the  greater  angle. 


In  the  triangle  ACB,  let  angle  ACB  be  greater  than 
angle  B. 

To  prove  AB  >  AC. 

Proof.  Now  AB  is  equal  to  AC,  or  less  than  AC,  or  greater 
than  AC. 

But  AB  is  not  equal  to  AC,  for  then  the  Z  C  would  be 
equal  to  the  Z  B,  §  154 

{being  A  opposite  equal  sides). 

And  AB  is  not  less  than  AC,  for  then  the  Z  C  would  be 
less  than  the  Z  ^,.  §158 

{if  two  sides  of  a  A  are  unequal,  the  A  opposite  are  unequal,  and  the 
greater  Z  is  opposite  the  greater  $ide). 

.-.  AB  is  greater  than  AC. 

aE.  D. 


Ex.  18.  ABC  a.B.d  ABD  are  two  triangles  on  the  same  base  AB,  and 
on  the  same  side  of  it,  the  vertex  of  each  triangle  being  without  the 
other,  li  AC  equal  AD,  show  that  BC  cannot  equal 
BD. 

Ex.  19.  The  sum  of  the  lines  which  join  a  point 
within  a  trianRln  to  the  tliroe  vertices  is  less  than 
the  perimeter,  but  greater  than  half  the  perimeter.    ^ 


62 


PLANE   GEOMETRY.  —  BOOK!  I. 


Peoposition  XXXIII.    Theoeem. 

160.  Two  triangles  are  equal  if  the  three  sides  of 
the  one  are  equal  respectively  to  the  three  sides  of 
the  other, 

B  B' 


In  the  triangles  ABC  and  A'BC,  letAB=^A'B',  AC^A'O, 
BG=B'C', 


To  prove 


A  ABC=  A  A'B'Q', 


Proof.  Place  A  A^B'C  in  the  position  AB^C,  having  its 
greatest  side  A'C  in  coincidence  with  its  equal  AC,  and  its 
vertex  at  J3',  opposite  B ;  and  draw  BB'. 

Since  AB  =  AB\  Hyp. 

A  ABB'  =  Z  AB'B,  §  154 

(in  an  isosceles  A  the  A  opposite  the  equal  sides  are  equal). 

Since  CB=CB\  '  Hyp. 

/.OBB'  =  ACB'B,  §154 

Hence,  Z  ABQ=  Z  AB'Q,  Ax.  2 

/.A  ABQ=  A  AB'0='A  A'B'C'  §  150 

{two  /^  are  equal  if  two  sides  and  included  Z  of  one  are  equal  to  two 
sides  and  included  Z  of  the  other). 


TRIANGLES. 


53 


Proposition  XXXIV.    Theorem. 

161.  Two  right  triangles  are  equal  if  a  side  and 
the  hypotenuse  of  the  one  are  equal  respectively  to  a 
side  and  the  hypotenuse  of  the  other. 


C    B' 


In  the  right  triangles  ABC  and  A^B'C,  let  AB=  A'B', 
and  AC  =  A' C. 


To  prove 


AABC=AA'B*C\ 


Proof.  Apply  the  A  ABC  to  the  A  A'B'C,  so  that  AB  shall 
coincide  with  A'B\  A  falling  upon  A\  B  upon  B\  and  C  and 
C"  upon  the  same  side  of  A'B*, 

Then  BC  will  take  the  direction  of  B'€\ 

{for  /L  ABC==  Z  A'B'C,  each  being  a  rt.  Z), 
Since  AC=A'0\ 

the  point  C  will  fall  upon  C\  §  121 

{two  equal  oblique  lines  from  a  point  in  a  ±  cut  off  equal  distances  from 
the  foot  of  the  ±). 


'.  the  two  A  coincide,  and  are  equal. 


Q.s.a 


54  PLANE   GEOMETRY.  —  BOOK  I. 


Proposition  XXXV.     Theorem. 

162.  Every  point  in  the  bisector  of  an  angle  is  equi- 
distant from  the  sides  of  the  an^le. 


Let  AD  be  the  bisector  of  the  angle  BAC,  and  let  0 
be  any  point  in  AD. 

To  prove  that  0  is  equidistant  from  AB  and  AC. 

Proof.   Draw  Oi^ and  OG  L  to  AB  and  -irrespectively. 

In  the  rt.  A  ^Oi^and  AOG 

AO=AO,  Iden. 

ZBAO  =  Z^AO,  Hyp. 

.\AAOF=AAOG,  §  148 

[two  rt.  ^  are  equal  if  the  hypotenuse  and  an  acute  Z  of  the  one  are  equal 
respectively  to  the  hypotenuse  and  an  acute  Z  of  the  other). 

...  0F=  OG, 

{homologous  sides  of  equal  ^). 
.'.  0  JA^quidistant  from  AB  and  AC. 


What  is  the  locils  of  a  point : 

Ex.  20.  At  a  given  distance  from  a  fixed  point  ?    §  57. 

Ex.  21.  Equidistant  from  two  fixed  points?    §  119. 

Ex.  22.  At  a  given  distance  from  a  fixed  straight  line  of  indefinite 
length  ? 

Ex.  23.  Equidistant  from  two  given  parallel  lines  ? 

Ex.  24.  Equidistant  from  tiie  extremities  of  a  given  line? 


TRIANGLES. 


55 


Proposition  XXXVI.    Theorem. 

163.  Every  -point  within  an  angle,  and  equidistant 
from,  its  sides,  is  in  tlie  bisector  of  the  angle. 


Let  0  be  equidistant  from  the  sides  of  the  angle 
BAC,  and  let  AO  join  the  vertex  A  and  the  point  0. 

To  prove  that  AO  is  the  bisdcior  of  Z  BAC. 

Proof.   Suppose  OF  and  OQ  drawn  X  to  AB  and  AG^ 

respectively. 


In  the  rt.  A  ^Oi^and  AOG 

0F=  00, 
AO  =  AO. 


Hyp. 

Iden. 


.'.A  AOF^AAOG,  §161 

{two  rt.  A  are  equalif  the  liypotenuse  and  a  side  oJ»^  one  arc  equal  to  the 
hypotenuse  and  a  side  of  the  otJ^r)/ 

.\ZFAO  =  ZGAO, 
{homologous  A  oj  equal  A). 

.'.  ^0  is  the  bisector  of  Z  BAC. 

aE.D. 

164.   Cor.    The  locus  of  a  point  within  an  angle,  and  equi- 
distant from  its  sides,  is  the  bisector  of  the  angle. 


66 


PLANE   GEOMETRY. —  BOOK   I. 


Quadrilaterals. 

165.  A  quadrilateral  is  a  portion  of  a  plane  bounded  by 
four  straight  lines. 

The  bounding  lines  are  the  sides,  the  angles  formed  by  these 
sides  are  the  angles,  and  the  vertices  of  these  angles  are  the 
ver^jc^rs,  of  the  quadrilateral. 

166.  A  trapezium  is  a  quadrilateral  which  has  no  two  sides 
parallel. 

167.  A  trapezoid  is  a  quadrilateral  which  has  two  sides,  and 

only  two  sides,  parallel. 

168.  ^A  parallelogram  is  a  quadrilateral  which  has  its  oppo- 
site sides  parallel. 


Trapezium. 


Trapezoid 


Parallelogram. 


169.  A  rectangle  is  a  parallelogram  which  has  its  angles 
right  angles. 

170.  A  rhomboid  is  a  parallelogram  which  has  its  angles 
oblique  angks. 

171.  A  square  is  a  rectangle  which  has  its  sides  equal. 

172.  A  rhombus  is  a  rhomboid  which  has  its  sides  equal. 


Square. 


Rectangle. 


Rhombus. 


Rhomboid. 


173.    The  side  upon  which  a  parallelogram  stands,  and  the 
opposite  side,  are  called  its  lower  and  upper  bases. 


QUADRILATERALS.  67 

174.  The  parallel  sides  of  a  trapezoid  are  called  its  bases^ 
the  other  two  sides  its  legs,  and  the  line  joining  the  middle 
points  of  the  legs  is  called  the  median. 

175.  A  trapezoid  is  called  an  isosceles  trapezoid  when  its 
legs  are  equal. 

176.  The  altitude  of  a  parallelogram  or  trapezoid  is  the 
perpendicular  distance  between  its  bases. 

177.  The  diagonal  of  a  quadrilateral  is 
straight  line  joining  two  opposite  vertices. 


Proposition  XXXVII.  •  Theorem. 

178.  Tlie  diagonal  of  a  parallelogram  divides  the 
figure  into  two  equal  triangles. 

C 


A  E 

Let  ABCE  be  a  parallelogram  and  AC  its  diagonal. 

To  prove  A  ABC=  A  AEC, 

In  the  A  ABC  and  AEC, 

AC=AC,  Iden. 

^ACB  =  ZCAE,    .  §104 

and  ZCAB  =  ^AC£!, 

{being  alt.-int.  AofW  lines.) 

.\AABC=AAECi  §147 

{having  a  side  and  two  adj.  A  of  the  one  equal  respectively  to  a  side  and 
two  adj.  A  of  the  other.) 

a  E.  D. 


68  PLANE  GEOMETRY.  —  BOOK   I. 


Proposition  XXXVIII.     Theorem. 

179.  In  a  parallelogram  the  opposite  sides  are  equalt 
and  the  opposite  angles  are  equal. 


Let  the  figure  ABCE  be  a.  parallelogram. 

To  prove  B0=  AU,  and  AB=  EC, 

also,  ZB  =  ZE,&ndZ  BAE  -= /.  BCE. 

Proof.  Draw  AC. 

AABO=AAEO,  §178 

{the  diagonal  of  a  O  divides  the  figure  into  two  equal  A). 

/.  BC=^  AE,  and  AB  =  CE, 
{being  homologous  sides  of  equal  A). 

Also,  Z  B  =  ZE,s.ndZBAE=ZBCE,  §112 

{having  their  sides  11  and  extending  in  opposite  directions  from 
their  vertices). 

Q.  E.  D. 

180.  Cor.  I  'arallel  lines  comprehended  between  parallel  lines 
are  equal.  A  B 

181.  Cor.  2.  Two  parallel  lines 
are    everywhere    equally   distant. 


For  if  AB  and  DC  are  parallel,  D  C 

Js  dropped  from  any  points  mAB  to  DC,  measure  the  distances 
of  these  points  from  DC.  But  thepe  Js  are  equal,  by  §  180; 
hence,  all  points  in  AB  are  equidistant  from  DC, 


QUADRILATERALS.  69 


Proposition  XXXIX.     Theorem. 

182.  If  two  sides  of  a  quadrilateral  are  equal  and 
parallel,  then  the  other  two  sides  are  equal  and  par- 
allel, and  the  figure  is  a  ■parallelogram. 


Let  the  figure  ABCE  be  a  quadrilateral,  having  the 
side  AE  equal  and  parallel  to  BG, 

To  prove  A B  equal  and  II  to  EC. 

Proof.  Draw  AC, 

In  the  A  ABC Q.ndi  AEC 

BC=^  AE,  Hyp. 

AC^AC,  Iden. 

ZBCA  =  ZCAE,  §104 

{being  alt.-int  A  qfW  lines). 

.'.AABC=AACE,  §150 

'(having  two  sides  and  the  included  Z  of  the  one  equal  respectively  to  two 
sides  and  the  included  Z  of  the  other). 

,\AB=EC, 

(being  homologous  sides  of  equal  A). 

Also,  ZBAC=ZACE, 

(being  homologous  A  of  equal  A). 

.-.  AB  is  II  to  EC, .  §  105 

(when  two  straight  lines  are  cut  by  a  third  straight  line,  if  the  alt.-int.  A 
are  equal,  the  lines  are  parallel). 

.'.  the  figure  ABCE  is  a  O,  §  168 

(the  opposite  sides  being  parallel),  a  e.  d. 


60  PLANE   GEOMETRY. — BOOK   I. 


Proposition  XL.     Theorem. 

183.   If  the  opposite  sides  of  a  quadrilateral  are 
equal,  the  figure  is  a  parallelogram. 


A 


Let  the  figure  ABCE  be  a  quadrilateral  having  BC=: 
AE  and  AB  =  EC. 

To  prove  figure  ABCE  a  O. 

Proof.  Draw  AQ, 

In  the  A  ABCM\d.  AEQ 

BC=  AE,  Hyp. 

AB=OE,  •   Hyp. 

AC=  AG.  Iden. 

.-.  A  ABC=  A  AEQ,  §  160 

{having  three  sides  of  the  one  equal  respectively  to  three  sides  of  the  other). 

.:ZACB  =  ZOAE, 

and  ZBAO=ZACE, 

{being  homologous  A  of  equal  A). 

.'.BClQ  W  to  AE, 

and  AB  is  II  to  EC,  §  105 

{when  two  straight  lines  lying  in  the  same  plane  are  cut  by  a  third  straight 

line,  if  the  alt.-mt.  A  are  equal,  the  lines  are  parallel). 

.-.  the  figure  ABCE  is  a  O,  §  168 

{having  its  opposite  sides  parallel). 


QUADRILATERALS.  61 

Proposition  XLI.     Theorem. 

184.   The  diagonals  of  a  parallelogram  bisect  each 
her. 


other. 


Let  the  figure  ADCE  be  a  parallelogram,  and  let 
the  diagonals  AC  and  BE  cut  each  other  at  0. 

To  prove  AO=OC,  and  BO  =  OK 

In  the  A  AOE  and  £00 

AE=BC,  §179 

{being  opposite  sides  of  a  O). 

ZOAE=ZOCB,  §104 

and  Z  OEA  =  Z  OBC, 

{being  alt.-int.  A  of  W  lines). 

.\  A  AOE  =:  A  BOO,  §147 

{having  a  side  and  two  adj.  A  of  the  one  equal  respectively  to  a  side  and 
two  adj.  A  of  the  other). 

.\AO=00,&naBO=OU, 
{being  homologous  sides  of  equal  i^). 


Ex.  25.  If  the  diagonals  of  a  quadrilateral  bisect  each  other,  the  figure 
is  a  parallelogram. 

Ex.  26.    The  diagonals  of  a  rectangle  are  equal. 

Ex.  27.   If  the  diagonals  of  a  parallelogram  are 
equal,  the  figure  is  a  rectangle.  -^ 

~  Ex.  28.   The  diagonals  of  a  rhombus  are  perpendicular  to  each  other, 
and  bisect  the  angles  of  the  rhombus. 

Ex.  29.   The  diagonals  of  a  square  are  perpendicular  to  each  other, 
ind  bisect  the  angles  of  the  square. 


62  PLANE   GEOMETRY. BOOK    I. 


Proposition  XLIL     Theorem. 

185.  Two  parallclo!irams,  having  two  sides  and  tlie 
included  arxgle  of  the  one  equal  respectively  to  two 
sides  and  the  zncluded  angle  of  the  other,  are  equal, 

B' g' 

7 


A'  D 

In  the  parallelograms  ABCD  and  AE'C'D',  let  AB^ 
A'B',  AD  =  A'D',  and  AA=AA\ 

To  'prove  that  the  UJ  are  equal. 

Apply  O  ABCD  to  O  A'WC'B\  so  that  AD  will  fall  on 
and  coincide  with  A^D\ 

Then  AB  will  fall  on  A'B\ 

{for  ZA==ZA',by  hyp.), 

and  the  point  B  will  fall  on  B\ 

(for  AB  =  A^B\  by  hyp.). 

Now,  BC  and  B'O^  are  both   II  to  AD'  and  are  drawn 

through  point  B'. 

.-.  the  lines  BC  and  B'C  coincide,  §  101 

and  (7  falls  on  B'C  or  B'C  produced. 

In  like  manner,  DC  and  D'C  are  II  to  A'B'  and  are  drawn 

through  the  point  D'. 

.'.  DC  and  D'C'  coincide.  §  101 

.*.  the  point  C  falls  on  D'C,  or  D'C  produced. 

.-.  (7  falls  on  both  B'C  and  D'C. 

.*.  Cmust  fall  on  the  point  common  to  both,  namely,  C. 

.'.  the  two  UJ  coincide,  and  are  equal. 

aE.  D 

186.   Cor.   Two  rectangles  having  equal  bases  and  altitudes 

are  equal. 


QUADRILATERALS.  63 


Proposition  XLIII.     Theorem. 

187.  //  three  or  more  parallels  intercept  equal  parts 
on  any  transversal,  they  intercept  equal  parts  on 
every  transversal. 


Let  the  parallels  AH,  BK,  CM,  DP  intercept  equal 
parts  BK,  KM,  MP  on  the  transversal  HP. 

To  prove  that  they  intercept  equal  parts  AB,  BCy  CD  on  the 

transversal  AD. 

Proof.   From  A,  B,  and  C suppose  AE,  BF,  and  CO  drawn 
II  to  HP, 

Then  AE=  HK,  BF^  KM,  CG  =  MP,        §  180 
[parallels  comprehended  between  paralleh  are  equal). 

,\AF=BF=CO.  Ax,  1 

Also  ZBAF=Z  CBF=  Z  BCG,  §  io6 

{being  ext.-int.  A  ofW  lines); 

and  .      ZAFB--=ZBFC=ZCGB,  §112 

(having  their  sides  li  and  directed  the  same  way  jrom  the  verticei). 

.-.  A  ABE=  A  BCF=  A  CDG,  §  147 

{each  having  a  side  and  two  ad}.  A  respectively  equal  to  a  side  and  two 
ad).  A  of  tht  others). 

,'.  AB  =  BC^  CD, 

{.hcmclogous  ndes  of  egital  A).  Q  ^-  °- 


64 


PLANE   GEOMETRY.  —  BOOK   I. 


188.  Cor.  1.  The  line  parallel  to  the  base  of  a  triangle  and 
bisecting  one  side,  bisects  the  other  side  cdso. 
For,  let  DE  Le  II  to  BC  and  bisect  AB. . 
Draw  through  A  a  Hne  II  to  BC.  Then 
this  line  is  II  to  DE,  by  §  111.  The  three 
parallels  by  hypothesis  intercept  equal 
parts  on  the  transversal  AB,  and  there-  -^ 
fore,  by  §187,  they  intercept  equal  parts  on  the  transversal 
AC \  that  is,  the  line  Z)^ bisects  AC. 

189.  Cor.  2.  The  line  which  joins  the  middle  points  of  two 
sides  of  a  triangle  is  parallel  to  the  third  side,  and  is  equal  to 
half  the  third  side.  For,  a  line  drawn  through  D,  the  middle 
point  of  AB,  II  to  BC,  passes  through  E,  the  middle  point  of 
^^>  l^y  §  1S8.  Therefore,  the  line  joining  Z)  and  E  coincides 
with  this  parallel  and  is  11  to  BC.  Also,  since  EF  drawn  II 
to  AB  bisects  AC,  it  bisects  BC,  by  §  188 ;  that  is,  BF=^  EC 
=  \  BC.  But  BEEF  is  a  O  by  construction,  and  therefore 
DE==BF=^BC. 

190.  Cor.  3.  The  line  which  is  parallel  to  the  bases  of  a  trap- 
ezoid and  bisects  one  leg  of  the  trap- 
ezoid bisects  the  other  leg  also.  For 
if  parallels  intercept  equal  parts  on 
any  transversal,  they  intercept  equal 
parts  on  every  transversal  by  §  187. 


i''\ 


191.  Cor.  4.  The  median  of  a 
trapezoid  is  parallel  to  the  bases,  and  is  equal  to  half  the  sum 
of  the  bases.  For,  draw  the  diagonal  EB.  In  the  A  AEB 
join  E,  the  middle  point  of  AE,  to  F,  the  middle  point  of  EB. 
Then,  by  §  189,  ^i^is  II  to  AB  and  --  ^AB.  In  the  A  EEC 
join  i^to  O,  the  middle  point  of  BC.  Then  FG  is  II  to  EC 
and  =^EC.  AB  and  FG,  being  II  to  EC,  are  II  to  each  other. 
But  only  one  line  can  be  drawn  through  F II  to  AB.  There- 
fore FG  is  the  prolongation  of  EF.  Hence  EFG  is  II  to  AB 
and  EC,  and  =  1(^^-1"  ^C'). 


EXERCISES. 


65 


Exercises. 

30,   The  bisectors  of  the  angles  of  a  triangle  meet  in  a  point  which  is 
equidistant  from  the  sides  of  the  triangle. 

lliST.  Let  the  bisectors  AD  and  BE  intersect  at  0. 
Then  0  being  in  AD  is  equidistant  from  AC  and  AB. 
(Why  ?)  And  0  being  in  BE  is  equidistant  from  BC 
and  AB.  Hence  0  is  equidistant  from  AC  and  BC, 
and  therefore  is  in  the  bisector  CF.     (Why  ?)  _ 

31.  The  perpendicular  bisectors  of  the  sides  of  a  triangle  meet  in  a 
point  which  is  equidistant  from  the  vertices  of  the 
triangle. 

Hint.   Let  the  J.  bisectors  EE^  and  DD^  intersect     D/ 
at  0.     Then  0  being  in  EE^  JB  equidistant  from  A  ^ 
and  C.    (Why  ?X   And  0  being  in  DD^  is  equidistant 
from  A  and  B.    Hence  0  is  equidistant  from  B  and  C,  and  therefore 
is  in  the  J.  bisector  FF^.  (Why  ?) 

32.  The  perpendiculars  from  the  vertices  of  a  triangle  to  the  opposite 
sides  meet  in  a  point.  a'-    -        4l.  »/ 

Hint.  Let  the  _fe.  be  AH,  BP,  and  CK:  '^"^  "*'"  ' 
Through  A,  B,  C  suppose  B'C,  A'C,  A'B' 
drawn  II  to  BC,  AC,  AB,  respectively.  Then 
AH  is  ±  to  B^C.  (Why?)  Now  ABCB^  and 
ACBC^  are  ZI7  (why?),  and'AB^^BC,  and  AC^ 
=  BC.  (Why  ?)  That  is,  A  is  the  middle  point  of  B^O^.  In  the  same  way, 
B  and  C  are  the  middle  points  of  A^C^  and  A^B^,  respectively.  There- 
fore, AH,  BP,  and  CiTare  the  _L  bisectors  of  the  sides  of  the  A  A^B^O^. 
Hence  they  meet  in  a  point.     (Why  ?) 

33.  The  medians  of  a  triangle  meet  in  a  point  which  is  two-thirds  of 
the  distance  from  each  vertex  to  the  middle  of  the  opposite  side. 

Hint.  Let  the  two  medians  AD  and  CE  meet  in  0. 
Take  i^the  middle  point  of  OA,  and  O  of  OC  Join 
GF,  FE,  ED,  and  DG.  In  A  AOC,  GF  ia  II  to  AC 
and  equal  to  .}  AC.  (Why  ?)  DE  is  II  to  AC  and  equal 
to  ^AC  (Why?)  Hence  DGFE  is  a  O.  (Why?) 
Hence  AF=^  FO  •-=  OD,  and  CG==GO=  OE.  (Why  ?)  ^ 
Hence,  any  median  cutS  off  on  any  other  median  two-thirds  of  the  dis- 
tance from  the  vertex  to  the  middle  of  the  opposite  side.  Therefore  the 
median  from  B  will  cut  off  AO,  two-thirds  of  AD;  that  is,  will  pass 
through  0 


ii\77 


H. 


66 


PLANE  GEOMETRY.  —  BOOK  I. 


Polygons  in  General. 

192.  A  polygon  is  a  plane  figure  bounded  by  straight  lines. 
The  bounding  lines  are  the  sides  of  the  polygon,  and  their 

sum  is  the  perimeter  of  the  polygon. 

The  angles  which  the  adjacent  sides  make  with  each  other 
are  the  angles  of  the  polygon,  and  their  vertices  are  the  ver- 
tices of  the  polygon. 

The  number  of  sides  of  a  polygon  is  evidently  equal  to  the 
number  of  its  angles. 

193.  A  diagonal  of  a  polygon  is  a  line  joining  the  vertices 
of  two  angles  not  adjacent ;  as  ^C,  Fig.  1. 

B 


Fig.  3. 


194.  An  equilateral  polygon  is  a  polygon  which  has  all  its 
sides  equal. 

195.  An  equiangular  polygon  is  a  polygon  which  has  all  its 
angles  equal. 

196.  A  convex  polygon  is  a  polygon  of  which  no  side,  when 
produced,  will  enter  the  surface  bounded  by  the  perimeter. 

197.  Each  angle  of  such*^  polygon  is  called  a  salient  angle, 
and  is  less  than  a  straight  angle. 

198.  A  concave  polygon  is  a  polygon  of  which  two  or  more 
sides,  when  produced,  will  enter  the  surface  bounded  by  the 
perimeter.     Fig.  3. 

199.  The  angle  FDE  is  called  a  re-entrant  angle,  and  is 
greater  than  a  straight  angle. 

If  the  term  polygon  is  used,  a  convex  polygon  is  meant. 


POLYGONS.  67 

200t  Two  polygQias  are  equal  when  they  eaji  be  divided  by- 
diagonals  into  the  same  number  of  triangles,  equal  each  to 
each,  and  similarly  placed  ;  for  the  polygons  can  be  applied 
to  each  other,  and  th©  corresponding  triangles  will  evidently 
coincide. 

201.  Two  polygons  are  muiuodl'^  equiangulw^  if  the  angles 
of  the  one  are  equal  to  the  angles  of  the  other,  each  to  each, 
when  tak«n  in  the  same  order.      Figs.  1  and  2. 

202.  The  eqaal  angles  in  mutually  equiangular  polygons 
are  called  homologous  angles ;  and  the  sides  which  lie  bekueen 
equal  angles  are  called  homologous  sides. 

203.  Two  polygons  are  mutually  equilateral,  if  the  sides  of 
the  one  are  equal  to  the  sides  of  the  others  each  to  each,  wli.en 
taken  in  the  same  order.     Figs.  1  and  2. 


Fio.  4  FiQ  5.  Fio.  6.  Fia.  7. 

Two  polygons  may  be  mutnially  equiangular  without  beiiig 
mutually  equilateral ;  as,  Figs.  4  and  5. 

And,  except  in  the  case  of  triangiles,  two  polygons  may  be 
mutuary  equilateral  without  being  mutually  equiangular  ;  as, 
Figs.  6  and  7. 

If  two  polygons  are  mutually  equilateral  and  equiangular, 
they  cure  equal,  for  they  may  be  applied  the  one  to  the  other 
so  as  to  coincide. 

204.  A  polygon  of  three  sides  is  called  a  trigon  or  triangle; 
one  of  four  sides,  a  tetragon  or  quadrilateral ;  one  of  five  sides, 
9.  pentagon ;  one  of  six  sides,  a  hexagon;  one  of  seven  sides,  a 
heptagon ;  one  of  eight  sides,  an  octagon :  one  oi  ten  sides,  a 
decagon;  one  of  twelve  sides,  a  dodecagoti. 


68 


PLANE    GEOMETRY. 


BOOK    I. 


Proposition  XLIV.     Theorem. 

205.  The  sum  of  the  interior  angles  of  a  -polygon  is 
equal  to  two  right  angles,  taken  as  many  times  less 
two  as  the  -figure  has  sides. 


Let  the  figure  ABODE F  be  a  polygon  having  n  sides. 

Te prove  ^  A-\-ZB-{-Z  C,  etc.  =  (n- 2)  2  rt.  A. 

Proof.    From  the  vertex  A  draw  the  diagonals  AC,  AD, 
and  AU. 

The  sum  of  the  A  of  the  A  =■  the  sum  of  the  A  of  the 
polygon. 


Now  there  ar€k(22_—  2)  A, 
and  the  sum  of  the  A  of  each  A  =  2  rt.  A. 


§138 


.-.  the  sum  of  the  zi  of  the  A,  that  is,  the  sum  of  the  A  of 
the  polygon  =  (n  -  2)  2  rt.  A.  ^  ^^  ^ 

206.    Cor.    The  sum  of  the  angles  of  a  quadrilateral  equals 

two  right  angles  taken  (4  —  2)  times,  i.e.,  equals  4  right  angles; 

and  if  the  angles  are  all  equal,  each  angle  is  a  right  angle.    In 

general,  each  angle  of  an  equiangular  polygon  of  n  sides  is 

2  (n  —  2)     ■ 
equal  to  "^ —  right  angles. 


POLYGONS.  69 


Proposition  XLV.    Theorem. 

207.  The  exterior  angles  of  a  polygon,  made  hy  -pro- 
ducing each  of  its  sides  in  succession,  are  together 
equal  to  four  right  angles. 


Let  the  figure  ABODE  be  a  polygon,  having  its  sides 
produced  in  succession. 

To  prove  the  sum  of  the  ext.  A  =  ^  rt.  A. 
Proof.  Denote  the  int.  A  of  the  polygon  hy  A^  B^  (7,  Z>,  E, 
and  the  ext.  Ahj  a,h,  c,  d,  e. 

ZA  +  Aa  =  2rt.A,  §90 

and  ZB  +  Zb  =  2  rt.  A, 

{being  sup.-adj.  A). 

In  like  manner  each  pair  of  adj.  A  =  2  rt.  A 

.'.  the  sum  of  the  interior  and  exterior  A  =  2rt.  A  taken 
as  many  times  as  the  figure  has  sides, 

or,  2  n  rt.  A. 

But  the  interior  ^  =f  2  rt.  ^  taken  as  many  times  as  the 
figure  has  sides  less  two,'  =  (n  —  2)  2  rt.  A, 

or,  -  2  n  rt.  ^  —  4  rt.  A. 

.'.  the  exterior  zi  =  4  rt.  A. 


70 


PLANE   GEOMETRY.  —  BOOK   I. 


Proposition  XLVI.     Theorem. 

208.  A  quadrilateral  which  has  two  adjacent  sides 
equal,  and  the  other  tivo  sides  equal,  is  symmetrical 
with  respect  to  the  diagonal  joining  the  vertices  of 
the  angles  formed  hy  the  equal  sides,  and  the  diago- 
nals intersect  at  right  angles. 


Let  ABCD  be  a  quadrilateral,  having  AB  =  AD,  and 
CB:=CD,  and  having  the  diagonals  AC  and  BD. 

To  prove  that  the  diagonal  A  C  is  an  axis  of  symmetry,  and 
is  S^to  the  diagonal  BD. 

Proof.   In  the  A  ^^(7  and  ^DC 

AB  =  AI),  and  BC=  BO,  Hyp. 

and  AC=Aa  Iden. 

.'.A  ABC  =- A  ABO,  §160 

{having  three  sides  of  the  one  equal  to  three  sides  of  the  other). 

.'.Z  BAC=rZBAO,  and  Z  BCA  -  Z  BOA, 
{homologous  A  of  equal  A). 

Hence,  if  ABC  is  turned  on  ^  C  as  an  axis,  AB  will  fall 
upon  AB,  CB  on  CB,  and  OB  on  OB. 

Hence  AC  is  an  axis  of  symmetry,  §  65,  and  is  J_  to  ,BD.^ 

Q.  E.  O. 


POLYGONS. 


71 


Pkoposition  XLVII.     Theorem. 

209.  If  a  figure  is  symmetrical  with  respect  to  two 
axes  perpendicular  to  each  other,  it  is  symmetrical 
with  respect  to  their  intersection  as  a  centre. 


B 

•    c 

i^--A 

.....:::^ 

A 

r' 

(\       -'" 

-'"^^ 

D 

X 

H 

0 

B 

^    G 

E 

Let   the  figure  ABODE FGH  be   symmetrica,!   with 
respect  to  the  two  axes  XX',  YY\  which  intersect  at  0. 

To  prove  0  the  centre  of  symmetry  of  the  figure. 

Proof.   Let  N  be  acy  point  m  the  perimeter  of  the  figure. 

Draw  NMIA.  to  YT,  end  IKL  J.  to  XX\ 

Join  LO,^OJV,  ^nd  KM. 

Now  KI=KL, 

{the  figure  being  symmetrical  with  respect  to  XX): 

But  KI=  OM, 

(lis  comprehended  between  lis  are  equal). 
.-.  KL  =  OM,  and  KLOM  is  a  O, 

{having  two  sides  eqvxil  and  parallel). 

.'.  LO  IS  equal  and  parallel  to  KM. 

In  like  manner  we  may  prove  OiV equal  and  parallel  to  KM. 

Hence  the  points  L,  0,  and  iVare  in  the  same  straight  line 

drawn  through  the  point  0  H  to  K3f;  and  L0=  ON,  since 

each  is  equal  to  KM. 

.'.  any  straight  line  LON",  drawn  through  0,  is  bisected  at  0. 

.'.  0  is  the  centre  of  symmetry  of  the  figure.         §  64 

a  E.  D. 


§61 
§180 
§182 
§179 


72  PLANE   GEOMETRY.  —  BOOK   I. 


Exercises. 

-  34.  The  median  from  the  vertex  to  the  base  of  an  isosceles  triangle  is 
perpendicular  to  the  base,  and  bisects  the  vertical  angle.  ] 

-.  35. .  State  and  prove  the  converse.  ' 

"'36.   The  bisector  of  an  exterior  angle  of  an  isosceles  triangle,  formed 
by  producing  one  of  the  legs  through  the  vertex,  is  parallel  to  the  base. 

"-37.  State  and  prove  the  converse.  \ 

^38.  The  altitudes  upon  the  legs  of  an  isosceles  triangle  are  equal. 

^  39.  State  and  prove  the  converse. 

^  40.  The  medians  drawn  to  the  legs  of  an  isosceles  triangle  are  equal. 

^41.  State  and  prove  the  converse.     (See  Ex.  33.) 

M2.  The  bisectors  of  the  base  angles  of  an  isosceles  triangle  are  equal. 

43.  State  the  converse  and  the  opposite  theorems. 

44.  The  perpendiculars  dropped  from  the  middle  point  of  the  base  of 
an  isosceles  triangle  upon  the  legs  are  equal. 

45.  State  and  prove  the  converse.  '^ 

'»  46.  If  one  of  the  legs  of  an  isosceles  triangle  is  produced  through  the 
vertex  by  its  own  length,  the  line  joining  the  end  of  the  leg  produced  to 
the  nearer  end  of  the  base  is  perpendicular  to  the  base. 

47.   Show  that  the  sum  of  the  interior  angles  of  a  hexagon  is  equal  to 
eight  right  angles. 

'  48.  Show  that  each  angle  of  an  equiangular  pentagon  is  f  of  a  right 
angle. 

"^  49.  How  many  sides  has  an  equiangular  polygon,  four  of  whose  dngles 
are  together  equal  to  seven  right  angles  ? 

50.   How  many  sides  has  a  polygon,  the  sum  of  whose  interior  angles 
is  equal  to  the  sum  of  its  exterior  angles  ? 

"•51.  How  many  sides  has  a  polygon,  the  sum  of  whose  interior  angles 
is  double  that  of  its  exterior  angles  ? 

^52.  How  many  sides  has  a  polygon,  the  sum^f  whose  exterior  anglee 
is  double  that  of  its  interior  angles  ? 


EXERCISES.  73 

^.    BAC  IS  a  triangle  having  the  angle  B  double  the  angle  A.    If  BD 
bisect  the  angle  B,  and  meet  AC  in  D,  show  that  BD  is  equal  to  AD. 

^54.  If  from  any  point  in  the  base  of  an  isosceles  triangle  parallels  to 
the  legs  are  drawn,  show  that  a  parallelogram  is  formed  whose  perimeter 
is  constant,  and  equal  to  the  sum  of  the  legs  of  the  triangle. 

^  65.  The  lines  joining  the  middle  point?  of  the  sides  of  a  triangle  divide 
the  triangle  into  four  equal  triangles. 

"^56.  The  lines  joining  the  miadle  points  of  the  side  of  a  square,  taken 
in  order,  enclose  a  square. 

""57.  The  lines  joining  the  middle  points  of  the  sides  of  a  rectangle 
(not  a  square),  taken  in  order,  enclose  a  rhombus. 

"•  58.  The  lines  joining  the  middle  points  of  the  sides  of  a  rhombus, 
taken  in  order,  enclose  a  rectangle. 

*'v59.  The  lines  joining  the  middle  points  of  the  sides  of  an  isosceles 
trapezoid,  taken  in  order,  enclose  a  rhombus  or  a  square. 

^  60.  The  lines  joining  the  middle  points  of  the  sides  of  any  quadri- 
lateral, taken  in  order,  enclose  a  parallelogram. 

^-  61.  The  median  of  a  trapezoid  passes  through  the  middle  points  of 
the  two  diagonals.  ' 

«  62.  The  line  joining  the  middle  points  of  the,diagonals  of  a  trapezoid 
18  equal  to  half  the  difference  oi  the  bases. 

"^    63.   In  an  isosceles  trapezoid  each  base  makes        > P 

equal  angles  with  the  legs.                                                /  \                 \ 
Hint.   Draw  CEW  DB,  /     \ \ 

"^64.   In  an  isosceles  trapezoid  the  opposite  angles    ^        ^  ^ 

are  supplementary. 

--  65.  If  the  angles  at  the  base  of  a  trapezoid  are  equal,  the  other ' 
angles  are  equal,  and  the  trapezoid  is  isasceles. 

n66.  The  diagonals  of  an  isosceles  trapezoid  are  equal. 

^  67.   If  the  diagonals  of  a  trapezoid  are  equal,  the  c D 

trapezoid  is  isosceles.  i^^\/'\ 

Hint.  Draw  CE  and  DF  JL  to  CD.    Show  that  ^  / ''/^J  \ 

ADF  and  BCE  are  equal,  that  ^  COD  and  AOB  are  jX         ]\\ 

iposceles,  and  that  i^  ^OC  and  i?0 7)  are  equal.  a   E           f  b 


74  PLANE    GEOMETRY. BOOK    I. 

'^di.  ABCD  is  a  pai-allel®grara,  E  and  F  the  middle  points  of  AD  and 
i? (7  respectively  :  show  that  BE  and  Di^will  trisect  the  diagonal  AC. 

"  69.  If  from  th«  diagonal  BD  of  a  square  ABCD,  BE  is  cut  off  equal 
to  BC,  and  EF  is  drawn  perpendicular  to  BD  to  meet  DO  at  F,  show 
that  DE  is  equal  to  EF,  and  also  to  FC 

•^  70.  The  bisector-  »f  the  vertical  angle  ^  of  a  triangle  ABC,  and  the 
bisectors  of  the  exterior  angles  at  the  base  formed  by  producing  the  sides 
AB  and  AC,  nieet  ki  a  point  which  is  equidistant  fr©m  the  base  and  the 
sides  produced. 

^71.  If  the  two  angles  at  the  base  of  a  triangle  are  bisect'ed,  aod 
through  the  point  of  meeting  of  the  bisectors  a  line  is  drawn  parallel  to 
the  base,  the  length  of  this  parallel  between  the  sides  is  equal  to  the  sum 
of  the  segments  of  the  sides  between  the  parallel  and  the  base. 

\  72.  If  one  of  the  acute  angles  of  a  right  triangle  is  double  the  otiier, 
the  hypotenuse  is  double  the  shortest  side. 

*~73.  The  sum  of  the  perpendiculars  dropped  from  any  point  in  the 
base  of  an  isoscelee  triangle  to  the  legs  is  constant,  o 

and  equal  to  tlie  altitude  upon  one  of  the  legs. 

Hint.   Let  PD  and  PE  be  the  two  Jfi,  BE  the 
altit-ude  upon  AC.     Draw  PO  JL  to  BF,  and  prove  /  ^;<?\i) 

the  ^PBG  and  PBD  equal.  zS^J^^S^ij 

■^  74.  Tlie  sum  of  the  perpendiculars  dropped  from  any  point  within  an 
equilateral  triangle  to  the  three  sides  is  constant,  and  equal  to  the 
altitude. 

Hint.  Draw  through  the  poin*  a  lin©  II  to  the  base,  arud  apply  Ex.  73, 

"^  75.  What  is  th«  kcus  of  all  points  equidistant  from  a  pasir  of  inter- 
secting lines  ? 

76.    In  the  triangle  CAB  the  bisector  of  the  angle  C  makes  with  the 
•perpendicular  from  Cto  AB  an  angle  equal  to  half  the  difference  of  the 
angles  A  and  B. 

-  77.  If  one  angle  of  an  isosceles  triangle  is  equal  to  60°,  the  triangle 
is  equilateral. 


^ 


BOOK  II. 
THE    CLRCTLE. 


Definitions. 

210.  A  cirde  is  a  portion  of  a  plane  bounded  by  a  cui'ved 
line  called  a  drcumference,  aU  points  of  which  are  e«[ually  dis- 
tant from  a  point  within  called  the  centre. 

211.  A  radkis  is  a  straight  line  drawn  from  the  centre  to  the 
circumference ;  and  a  diameie)'  is  a  straight  line  drawn  through 
the  centre,  having  its  extremities  in  the  circumference. 

By  the  definition  of  a  circle,  all  its  radii  are  eqnal.  All  its 
diametere  are  equal,  since  the  diameter  is  equal  to  two  radii. 

212.  A  secant  is  a  straight  line  which  intersects  the  circum- 
ference in  two  points  ;  as,  AD,  Fig.  1. 

213.  A  tangent  is  a  straight  line  which  touches  the  circum- 
ference but  does  not  intersect  it;    as, 

BC,  Fig.  1.  The  point  in  which  the 
tangent  touches  the  circumference  is 
called  the  point  of  contact,  or  'po'mi  of 
tangency. 

214.  Two  drcumfet^ences  are  tangent 
to  e^oh  other  when  they  are  both  tan-  ^^^-  ^' 

gent  to  a  straight  line  a-t  the  same  point;  and  are  tangent 
internally  or  externally,  according  as  one  circumference  lies 
wholly  withm  or  without  the  other. 


76  PLANE   GEOMETRY. —  BOOK    IL 

215.  An  arc  of  a  circle  is  any  portion  of  the  circumference. 
An  arc  equal  to  one-half  the  circumference  is  called  a  semi' 
circumference. 

216.  A  chord  is  a  straight  line  having  its  extremities  in  the 
circumference. 

Every  chord  subtends  two  arcs  whose  sum  is  the  circum- 
ference; thus,  the  chord  AB  (Fig.  3)  subtends  the  smaller  arc 
AB  and  the  larger  arc  BCDEA.  If  a  chord  and  its  arc  are 
spoken  of,  the  less  arc  is  meant  unless  it  is  otherwise  stated. 


217.  A  segment  of  a  circle  is  a  portion  of  a  circle  bounded 
by  an  arc  and  its  chord. 

A  segment  equal  to  one-half  the  circle  is  called  a  semicircle. 

218.  A  sector  of  a  circle  is  a  portion  of  the  circle  bounded 
by  two  radii  and  the  arc  which  they  intercept. 

A  sector  equal  to  one-fourth  of  the  circle  is  called  a  quadrant. 

219.  A  straight  line  is  inscribed  in  a  circle  if  it  is  a  chord. 

220.  An  angle  is  inscribed  in  a  circle  if  its  vertex  is  in  the 
circumference  and  its  sides  are  chords. 

221.  An  angle  is  inscribed  in  a,  segment  if  its  vertex  is  on 
the  arc  of  the  segment  and  its  sides  pass  through  the  extrem- 
ities of  the  arc. 

222.  A  polygon  is  inscribed  in  a  circle  if  its  sides  are 
chords  of  the  circle. 

223.  A  circle  is  inscribed  in  a  polygon  if  the  circumference 
touches  the  sides  of  the  polygon  but  does  not  intersect  them. 


ARCS   AND   CHORDS.  77 

224.  A  polygon  is  circumscribed  about  a  circle  if  all  the 
sides  of  the  polygon  are  tangents  to  the  circle. 

225.  A  circle  is  circumscnbed  about  a  polygon  if  the  circum- 
ference passes  through  all  the  vertices  of  the  polygon. 

226.  Two  circles  are  equal  if  they  have  equal  radii ;  for 
they  will  coincide  if  one  is  applied  to  the  other ;  conversely, 
two  equ^l  circles  have  equal  radii. 

Two  circles  are  concentric  if  they  have  the  same  centre. 

Proposition  I.     Theorem. 

227.  The  diameter  of  a  circle  i's  greater  than  any 
oilier  chord;  and  bisects  the  circle  and  the  circum- 
ference. 


P 
Let  AB  be  the  diameter  of  the  circle  AM  DP,  and 
AE  any  other  chord. 

To  prove  AB  >  AE,  and  that  AB  bisects  the  circle  and  the 
circuviference. 

Proof,    I.    From  C,  the  centre  of  the  O,  draw  CE. 
CE^CB, 
•  {being  radii  of  the  same  circle). 

But  AC-\-CE>  AE,  §137 

{the  sum  of  two  sides  of  a  A  is  "^  the  third  side). 

Then  AC+CB>  AE,  or  AB  >  AE.  Ax.  9 

II.    Fold  over  the  segment  A  MB  on  AB  as  an  axis  until  it 

falls  upon  ABB,  §  59.    The  points  A  and  B  will  remain  fixed; 

therefore  the  arc  A  MB  will  coincide  with  the  arc  ABB ; 

because    all    points   in    each    are    equally  distant   from    the 

centre  C.  §  210 

Hence  the  two  figures  coincide  throughout  and  are  equal.  §  59 

Q.E.  D. 


78  PLANE    GEOMETRY.  —  BOOK    II. 


Proposition  II.     Theorem. 

228.   A  straight  line  cannot  intersect  the  cirewm- 
ference  of  a  circle  in  more  than  two  points. 


Let  HK  be  any  line  cutting  the  circumference  AMP, 

To  prove  that  UK  can  intersect  the  circumference  in  only  two 
points. 

Proof.  If  possible,  let  UK  intersect  the  circumference  in 
three  points  H,  P,  and  K. 

From  0,  the  centre  of  the  O,  draw  OH,  OP,  and  OK 

Then  OS,  OF,  and  OJTare  equal, 

{being  radii  of  the  same  circle). 

Hence,  we  have  three  equal  straight  lines  OIT,  OF,  and  OK 
drawn  from  the  same  point  to  a  given  straight  line.  But  this 
is  impossible,  §  120 

{only  two  equal  straight  lines  can  he  d^rawn  from  a  point  to  a  straight  line). 

Therefore,  KK  can  intersect  the  circumference  in  only  two 
points.  Q.  E.  a 


ARCS   AND   CHORDS.  79 


Proposition  III.    Theorem. 

229.  In  the  same  circle,  or  equal  circles,  equal  an- 
gles at  the  centre  intercept  equal  arcs;  conversely, 
equal  arcs  subtend  equal  angles  at  the  centre. 


p  P' 

In  the  equal  circles  ABP  and  A'B'P'  let  ^0=^/.O. 

To  prove  arc  R8  =  arc  E'S*. 

Proof.  Apply  O  ABP  to  O  A'B'P, 

so  that  Z  0  shall  coincide  with  Z  (7. 

a  will  fall  upon  B',  and  3  upon  S\  §  226 

(for  0R=  0^R\  and  OS^  0^8^,  being  radii  of  equal  ©). 

Then  the  arc  B8  will  coincide  with  the  arc  B'S\ 

since  all  points  in  the  arcs  are  equidistant  from  the  centre. 

§210 
.-.  arc  i?.S'=arc  B'8'. 

Conversely  :      Let  arc  RS=  arc  R'S^. 

To  prove  Z.O  =  Z.a. 

Proof.  Apply  O  ^^Pto  O  A'B'F\  so  that  arc  BS  shall  fall 
upon  arc  B'jS',  R  falling  upon  B\  8  upon  8\  and  0  upon  0'. 

Then  RO  will  coincide  with  i?'6>',  and  80  with  aS"0'. 

.".  A  O  and  0'  coincide  and  are  equal.  q.  e.  d. 


80  PLANE   GEOMETRY. — -BOOK   II. 


Proposition  IV.   Theorem. 

230.  In  the  same  circle,  or  equal  circles,  if  two 
chords  are  equal,  the  arcs  which  they  subtend  are 
equal;  conversely,  if  two  arcs  are  equal,  the  chords 
which  subtend  them  are  equal. 


p  P' 

In  the  equal  circles  ABP  and  A'B'P',  let  chord  RS  = 
chord  R'S'. 

To  prove  arc  US—  arc  B'S'. 

Proof.       Draw  the  radii  OB,  OS,  O'li',  and  O'S'. 

In  the  A  OBS  smd  O'B'S' 

IiS=  E'S',  Hyp. 

the  radii  OB  and  OS^  the  radii  O'B'  and  O'S'.   §  226 

.'.ABOS=AB'0'S',  §160 

{three  sides  of  the  one  being  equal  to  three  sides  of  the  other). 

.'.ZO^-=ZO', 

.'.  &VC  BS=  arc  B'S',  §229 

(in  equal  ®,  eqiuil  A  at  the  centre  intercept  equal  arcs). 

aE.o. 
Conversely  :     Let  arc  RS  =  arc  R^S'. 

To  prove  '  chord  BS=  chord  B'S'. 

Proof."  ZO^ZO',  §  229 

(equal  arcs  in  equal  ©  subtend  equal  A  at  the  centre), 

and  OB  and  0S=  O'B'  and  O'S',  respectively.    §  226 

.■.AOBS=AO'B'S',  §150 

(having  two  sides  equal  each  to  each  and  the  included  A  equal). 

.-.  chord  BS=  chord  B'S'.  q.e.d. 


ARCS   AND    CHORDS.  81 


Proposition  V.     Theorem. 


231.  In  the  same  circle,  or  equal  circles,  if  tivo  arcs 
are  unequal,  and  eaxih  is  less  than  a  semi-circumfer- 
ence, the  greater  arc  is  subtended  hy  the  greater 
chord;  conversely,  the  greater  chord  subtends  the 
greater  arc. 


In  the  circle  whose  centre  is  0,  let  the  arc  AMB  he 
greater  than  the  arc  AMF. 

To  prove      chord  AB  greater  than  chord  AF. 
Proof.  Draw  the  radii  OA,  OF,  and  OB. 

Since  F  is  between  A  and  B,  OF  will  fall  between  OA  and 
OB,  and  Z  AOB  be  greater  than  Z  AOF. 
Hence,  in  the  A  AOB  and  AOF, 

the  radii  OA  and  OB  =  the  radii  OA  and  OF, 
but  Z  ^0^  is  greater  than  Z  AOF. 

.\AB>AF,  §152 

{the  ^  having  two  sides  equal  each  to  each,  bid  the  included  A  unequal). 
Conversely:  Let  AB  be  greater  than  AF.        'I    / 

To  prove         arc  AB  greater  than  arc  AF.         t    '   ,• 
In  the  A  AOB  and  AOF,  '*  ) 

OA  and  0B=  OA  and  Oi^  respectively.   ;.    '  • 
But  AB  is  greater  than  AF.  "'      Hyp. 

.-.  Z  ^0^  is  greater  than  Z.  AOF,  §  153 

(i/i€  A  having  two  sides  equal  each  to  each,  but  the  third  sides  unequal). 
.-.  OB  falls  without  OF. 
.•.  arc  AB  is  greater  than  arc  AF.  q  e. d. 


82 


PLAJTE    GEOMETRY.  —  -ROOK    II. 


Proposition  VI.     Theorem. 

232.   The  radius  perpendicular  to  a  cJwrd  bisects 
the  chard  and  the  arc  subtended  by  it. 

E 


Let  AB  be  the  chord,  and  let  the  radius  OS  be  per- 
pendicular to  AB  at  M. 

To  prove    AM—  BM,  and  are  AS  =  arc  BS. 

Proof.   Draw  OA  and  OB  from  0,  the  centre  of  the  circle. 

In  the  rt.  A  0AM md  OBM 

the  radius  OA  —  the  radios  OB, 

and  0M=  OM.  Iden. 

.'.A0AM=^/10BM,  §161 

{having  the  hypotenu»e  and  a  side  of  one  equal  to  the  hypotermse  and  a 
side  of  the  other). 

.-.  AM=  BM, 
Sind  Z  AOS  =  Z  B08. 
.'.  arc  AS—  arc  BS, 

{equal  A  at  the  centre  intercept  equal  arcs  on  the  eireumfer&nce). 

aE.D. 

233.  Cor.  1.  Tke  perpendicular  erected  at  the  middle  of  a 
chord  passes  through  the  centre  of  the  circle.  For  the  centre  is 
equidistant  from  the  extremities  of  a  chord,  and  is  therefore  in 
the  perpendicular  erected  at  the  middle  of  the  chord.      §  122 

234.  Cor.  2.  The  perpendicular  erected  at  tke  middle  of  a 
chord  bisects  the  arcs  of  the  chord. 

235.  Cor.  3.  The  locus  of  the  middle  pemts  of  a  system  of 
parallel  chords  is  the  dia7ncter  perpendicular  to  thein. 


ARCS    AND    CHORDS.  .       .,  83 


Proposition  VII.     Theorem. 

236.   In  the  same    circle,   or    equal    circles,   equal 
chords    are  equally  distant   from  tJie  centre ;    and 

CONVERSELY. 


Let  AB  and  CF  be  equal  chords  of  the  circle  ABFC. 

To  prove     AB  ctnd  CF eqiddisUxniJroin  the  centre  0. 
Proof.  Draw  OPl.  to  AB,  OS  A.  to  CF,  and  join  OA  and  OC. 

OP  and  OH  bisect  AB  and  CF,  §  232 

(a  radius  A.  to  a  chord  bisects  it). 

Hence,  in  the  rt.  A  OF  A  and  OHC 

AP^CH,  Ax.  7 

the  radius  OA  =  the  radius  OC. 

.\AOPA  =  AOIK\  §161 

{having  a  side  and  hypotenuse  of  the  one  equal  to  a  side  enid  hypotenuse 
of  the  other). 

.'.  OP  =011. 

.*.  ^^  and  CF  'Ave  equidistant  from  O. 

Conversely  :   Let  OP  =  OH. 

To  prove  AB  =  CF. 

Proof.    In  the  rt.  A  OPA  and  OHC 

the  radius  OA  =  the  radius  OC,  and  0P=-  OH{hj  hyp.). 

.-.  A  OPA  and  OHC  sue  equal.  §  161 

..AP=CH. 

,',AB^CF.  Ax.  6. 

%  s<  o< 


84  PLANE   GEOMETRY.  —  BOOK    II. 


Proposition  VIII.     Theorem. 

237.  In  the  same  circle,  or  equal  circles,  if  two 
chords  are  unequal,  they  are  unequally  distant  from 
the  centre,  and  the  greater  is  at  the  less  distance. 


In  the  circle  whose  centre  is  0,  let  the  chords  AB 
and  CD  be  unequal,  and  AB  the  greater;  and  let  OE 
and  OF  be  perpendicular  to  AB  and  CD  respectively. 

To  prove  OE  <  OF. 

Proof.    Suppose  AG  drawn  equal  to  CD,  and  Oil  1.  to  AG. 

Then  OII=r  OF,  §  236 

{in  the  same  O  two  equal  chords  are  equidistant  from  the  centre). 

Join  FH. 

OF  and  Off  bisect  AB  and  AG,  respectively,      §  232 
(a  radius  L  to  a  chord  bisects  it). 

Since,  by  hypothesis,  ^^  is  greater  than  CD  or  its  equal  A  G, 
AF,  the  half  of  AD,  is  greater  than  Aff,  the  half  of  AG. 

.*.  the  Z  AffF  is  greater  than  the  Z  AFIl      §  158 
{tJie  greater  of  two  sides  of  a  A  has  the  greater  Z  opposite  to  it). 

Therefore,  the  Z  OlIF,  the  complement  of  the  Z  AffF,  is 
less  than  the  Z  OEff,  the  complement  of  the  Z  AFff. 

.\OF<Off,  §159 

{the  greater  of  two  A  of  a  A  has  the  greater  side  opposite  to  it). 

,'.  OF  <  OF,  the  equal  of  Off. 

Q.  e  o 


ARCS   AND   CHORDS.  85 


Proposition  IX.     Theorem. 

238.  Conversely  :  In  the  same  circle,  or  equal  cir- 
cles, if  two  chords  are  unequally  distant  from  the 
centre,  they  are  unequal,  and  the  chord  at  the  less 
distance  is  the  greater. 


In  the  circle  whose  centre  is  0,  let  AB  and  CD  be 
andqually  distant  from  0;  and  let  OK  perpendicular 
to  AB  be  less  than  OF  perpendicular  to  CD. 

To  prove  AB  >  CD. 

Proof.    Suppose  AO  drawn  equal  to  CD,  and  OH  1.  to  A  G. 

Then  011=  OF,  §  236 

{in  the  same  O  two  equal  chords  are  equidistant  from  the  centre). 

Hence,  OE  <  OH, 

Join  EH 

In  the  A  OEHWiQ  Z  OHE  is  less  than  the  Z  OEH,  §  158 
{the  greater  of  two  sides  of  a  A  has  the  greater  Z  opposite  to  it). 

Therefore,  the  Z  AHE,  the  complement  of  the  Z  OHE,  is 
greater  than  the  Z  A  EH,  the  complement  of  the  Z  OEH 

:.AE>AH,  §159 

{the  greater  of  two  A  of  a  A  has  the  greater  side  opposite  to  it). 

But  AE=iAB,  and  AH=iAG. 
,'.An>  .1(7;  hence  AB>  CD,  the  equal  of  AG. 


86  PLANE    GEOMETRY.    —BOOK    TI. 


Proposition  X.     Theorem. 

239.   A  straight  line  perpendicular  to  a  radius  at 
its  extremity  is  a  tangent  to  the  circle. 


II  A  -^ 

Let  MB  be  perpendicular  to  the  radius  OA  at  A, 

To  prove  MB  tangent  to  the  circle. 

Proof.    From  0  draw  any  other  line  to  MB,  as  OCR. 

OR>OA,  §114 

(a  ±  is  the  shortest  line  from  a  point  to  a  straight  line). 

.'.  the  point  ^is  without  the  circle. 
Hence,  every  point,  except  A,  of  the  line  MB  is  without  the 
circle,  and  therefore  MB  is  a  tangent  to  the  circle  at  A.  §  213 

a  E.  D. 

240.  Cor.  1.  A  tangent  to  a  circle  is  perpendicular  to  the 
radius  drawn  to  the  point  of  contact.  For,  if  MB  is  tangent 
to  the  circle  at  A,  every  point  of  MB,  except  A,  is  without 
the  circle.  -Hence,  OA  is  the  shortest  line  from  0  to  MB,  and 
is  therefore  perpendicular  to  MB  (§  114)  ;  that  is,  MB  is  per- 
pendicular to  OA. 

241.  Cor.  2.  A  perpendicular  to  a  tangent  at  the  point  of 
contact  passes  through  the  centre  of  the  circle.  For  a  radius  is 
perpendicular  to  a  tangent  at  the  point  of  contact,  and  there- 
fore, by  §  89,  a  perpendicular  erected  at  the  point  of  contact 
coincides  with  this  radius  and  passes  through  the  centre. 

242.  Cor.  3.  A  perpendicular  let  fall  from  the  centre  of  a 
circle  upon  a  tangent  to  the  circle  pafises  through  the  point  of 
contact 


ARCS   AND    CHORDS. 


87 


Proposition  XI.     Theorem. 

243.  Parallels    intercept   equal   arcs   on   a  circuni- 
ference. 


Fig.  2. 
Let  AB  and  CD  be  the  two  parallels. 

Case  I.     When  AB  is  a  tanf/ent,  cmd  CD  a  svcatd.     Fig.  1. 

Suppose  AB  touches  the  circle  at  F. 
To  prove  arc  CF=  arc  DF. 

Proof.  Suppose  FF^  drawn  _L  to  AB. 

This  J_  to  AB  at  i^  is  a  diameter  of  the  circle.     §  241 
It  is  also  ±  to  CD.  §  102 

.-.  arc  CF-=  arc  DF,  §  232 

(a  radius  ±  to  a  chord  bisects  the  chord  and  its  subtended  arc). 

Also,  arc  FCF'  =  arc  FDF\  §  227 

.-.  arc  (FCF'  -  FC)  ■=  arc  (FDF'  -  FD),       §    82 


arc 


CF' 


DF' 


that  is, 

Case  II.     When  AB  and  CD  are  secants.     Fig.  2. 
Suppose  ^i'^  drawn  11  to  CD  and  tangent  to  the  circle  at  J/. 

Then  arc  AM  —  arc  BIT 

and  arc  C3I    =  arc  DM  Case  I. 


.•.by  subtraction,  arc  AC    =  arc  BD 

Case  III.     When  AB  and  CD  are  tangents.     Fig.  3. 
Suppose  AB  tangent  at  E,  CD  at  F,  and  GH 11  to  AB. 

Then  arc  OF    =  arc  FR  Case  I. 

and  arc  GF    =  arc  JIF 

.".  bv  addition,       arc  EGF=  arc  EHF      '  q.  e.  d. 


88  PLANE   GEOMETRY.  —  BOOK    II. 


Proposition  XII.     Theorem. 

244.    Through  three  points  not  in  a  straight  line, 
one  cireumference,  and  only  one,  ean  he  drawn. 


Let  A,  B,  G  be  three  points  not  in  a  straight  line. 

To  prove  that  a  circumference  can  be  drawn  throiujlt  A,  B, 
and  (7,  and  only  one. 

Proof.  Join  v4i5and  ^C. 

At  the  middle  points  ^i^AB  and  .SCstippose  Js  erected. 

Since  BC\^  not  the  prelongation  of  AB,  these  Js  will  inter- 
sect in  some  point  0. 

The  point  0,  being  in  the  J_  to  ^^  at  its  mi^le  point,  is 

equidistant  from  A  and  B\  and  being  in  the  J_  to  BC 2X  its 

middle  point,  is  equidistant  from  B  and  C,  §  12:^ 

{every  point  in  the  perpendicular  bisector  of  a  straight  line  is  equidista  it 

from  the  extremities  of  the  straight  line). 

Therefore  0  is  equidistant  ivom/  A,  B,  and  C\  and  a  cir- 
cumference described  from  0  as  a  centre,  and  with  a  radiu^ 
OA,  will  pass  through  the  three  given  points. 

Only  one  circumference  can  be  made  to  pass  through 
these  points.  For  the  centre  of  a  circumference  passing 
through  the  three  points  must  be  in  both  perpendiculars,  and 
hence  at  their  intersection.  As  two  straight  lines  can  intei*- 
Rect  in  only  one  point,  0  is  the  centre  of  the  only  circumfer- 
ence that  can  pass  through  the  three  given  points.  ;i.  e.  d. 

245.  Cor.  Two  circumferences  can  intersect  in  only  two 
points.  For,  if  ^wo  circumferences  Imve  three  points  common, 
they  coincide  and  form  one  circumference. 


Tangents.  89 

Proposition  XT  IT.     Theorem. 

246.  The  tangervts  to  a  circle  drawn  fronv  an  exte- 
rior point  are  equal,  and  make  equal  angles  with, 
the  line  joining  the  point  to  th£^  centre. 


<: 

Let  AB  and  AC  be  tangents  from  A  to  the  circle 
whose  centre  is  0,  and  AO  the  line  joining  A  to  0. 

To  prove     AB  =  AC,  and  ZBAO  =  ZCAO. 
Proof.  Draw  0£  and  OC. 

AB  is  X  to  OB,  and  AC ±  to  OC,  §  240 

(a  tangent  to  a  circle,  is  A.  to  the  radius  drawn  to  the  jwiiit  of  contact). 
In  the  rt.  A  0^^  and  OAC 

OB^rOa 
{radii  of  the  same  circle). 

OA  =  OA.  Iden. 

.\AOAB  =  AOAC,  U61 

{having  a  side  and  hypotenuse  of  the  one  equal  to  a  side  and  hypotenme 
of  the  other). 

.'.AB^-AC, 
and  ZBAO^ZCAO.  '      q.e.d. 

247.  Def.  The  line  joining  the  centres  of  two  circles  is 
called  tiio  luie  of  centres. 

248.  Def.  A  common  tangent  to  two  circles  is  called  a 
common  cxtei'ior  imige^it  when  it  does  not  Cut  the  line  of  cen- 
tres, and  a  coTnmon  interior  tangent  when  it  cuts  the  line  of 
centres. 


90  PLANE    GEOMETEY. BOOK    II. 


Proposition  XIV.     Theorem. 

249.  If  two  circumferences  intersect  each  others  the 
lijie  of  centres  is  perpendicular  to  their  coTnmon 
chord  at  its  middle  point. 


Let  C  and  C  be  the  centres  of  two  circumferences 
which  intersect  at  A  and  B.  Let  AB  be  their  common 
chord,  and  CC  Join  their  centres. 

To  prove        CC^  1.  to  AB  at  its  middle  point. 

Proof.  A  J_  drawn  through  the  middle  of  the  chord  AB 
passes  through  the  centres  C  and  C\  §  233 

(«  ±  erected  at  the  middle  of  a  chord  passes  through  the  centre  of  the  O). 

.'.  the  line  CC,  having  two  points  in  common  with  this  J_, 
must  coincide  with  it. 

.'.  CC  is  JL  to  AB  at  its  middle  point.  q.  e.  d. 


Ex.  78.   Describe  the  relative  position  of  two  circles  if  the  line  of 
centres : 

(i.)  is  greater  than  the  sum  of  the  radii ; 
(ii.)  is  equal  to  the  sum  of  the  radii ; 

(iii.)  is  less  than  the  sum  but  greater  than  the  difference  of  the  radii ; 
(iv.)  is  equal  to  the  difference  of  the  radii ; 
(v.)  is  less  than  the  diflference  of  the  radii. 
Illustrate  each  case  by  a  figure. 


TANGENTS.  9l 


Proposition  XV.     Theorem. 

250.   If  two  circumferences  are  tangent  to  each  otioer, 
tlie  line  of  centres  passes  through  the  point  of  contact. 


Let  the  tvro  circumferences,  'whose  centres  are  C 
and  C,  touch  each  other  at  O,  in  the  straight  line  AB, 
and  let  GC  be  the  straight  line  joining  their  centres. 

To  prove  0  is  in  the  straight  line  CC 

Proof.  A  ±  to  AB,  drawn  through  the  point  0,  passes 
through  the  centres  C  and  C",  §  241 

{a  ±  to  a  tangent  at  the  point  of  contact  passes  through  the  centre 
of  the  circle). 

.-.  the  line  CC\  having  two  points  in  common  with  this  J, 
must  coincide  with  it. 

.'.  0  is  in  the  straight  line  CC.  ae-o. 


Ex.  79.   The  line  joining  the  centre  of  a  circle  to  the  middle  of  £■ 
chord  is  perpendicular  to  the  chord. 

Ex.  80.   The  tangents  drawn  through  the  extremities  of  a  diameter 
are  parallel. 

xEx.  81.   The  perimeter  of  an  inscribed  equilateral  triangle  is  equal 
io  half  the  perimeter  of  the  circumscribed  equilateral  triangle. 

y  Ex.  82.   The  sum  of  two  opposite  sides  of  a  circumscribed  quadri- 
lateral is  equal  to  the  sum  of  the  other  two  sides. 


92  PLANE    GEOMETRY.  —  BOOK    II. 


Measurement. 

251.  To  7)ieasure  a  quantity  oi'  any  kind  is  to  find  how  many 
times  it  contains  another  known  quantity  of  the  same  kind. 

Thus,  to  measure  a  line  is  to  find  how  man}^  times  it  con- 
tains another  known  line,  called  the  linear  unit. 

The  number  which  expresses  how  many  times  a  quantity 
contains  the  unit,  joined  with  the  name  of  the  unit,  is  called 
the  numerical  measure  of  that  quantity  ;  as,  5  yards,  etc. 

252.  The  magnitude  of  a  quantity  is  always  relative  to  the 
magnitude  of  another  quantity  of  the  same  Ichid.  No  quantity 
is  great  or  small  except  by  comparison.  This  relative  magni- 
tude is  called  their  ratio,  and  is  expressed  by  the  indicated 
quotient  of  their  numerical  measures  when  the  same  unit  of 
measure  is  applied  to  both. 

The  ratio  of  a  to  5  is  written  -,  o\  a:h.  * 

0 

253.  Two  quantities  that  can  be  expressed  in  integers  in 
terms  of  a  common  unit  are  said  to  be  eommerisurablc.  The 
common  unit  is  called  a  common  'measure,  and  each  quantity 
is  called  a  multiple  of  this  common^measure.^ 

Thus,  a  common  measure  of  1\  feet  and  3|-  feet  is  -J-  of  a 
foot,  which  is  contained  15  times  in  2\  feet,  and  22  times  in 
3-|  feet.  Hence,  2-^  feet  and  3f  feet  are  multiples  of  -J-  of  a 
foot,  2^  feet  being  obtained  by  taking  \  of  a  foot  15  times,  and 
3|  by  taking  -J  of  a  foot  22  times. 

254.  When  two  quantities  are  incomrncnsurabl^  th^t  is, 
have  no  common  unit  in  terms  of  which  both  quantities  can  be 
expressed  in  integers,  it  is  impossible  to  find  a  fraction  that 
will  indicate  the  exact  value  of  the  ratio  of  the  given  quanti- 
ties. It  is  possible,  however,  by  taking  the  unit  sufliciently 
small,  to  find  a  fraction  that  shall  difiPer  from  the  true  value 
of  the  ratio  by  as  little  as  we  please. 


KATIO.  U3 

Thus,  suppose  a  uiid  h  to  denote  two  lines,  such  tliaF 
a a         ,- 

I. --       ,      r^^- 

Now  Vil--  1.414^350...,..,  a  value  greater  than  1.414^13^ 
but  less  than  1.41^214^ 

If,  then,  a  viilliorith  part  of  h  be  taken  as  the  unit,  the  value 

of  the  ratio  -  lies  between  IvoHiff  ^^"^  fHlftv'  ^"^^  there- 
fore differs  from  either  of  these  fractions  by  less  than  y^T^iTrcir- 

By  carrying  the  decimal  farther,  a  fraction  may  be  found 
that  will  differ  from  the  true  value  of  the  ratio  by  less  than  a 
billionth,  a  trilUonth,  or  any  other  as^si'jne.d  value  whatever. 

Expressed  generally,  when  a  and  h  are  incommensurable, 
and  b  is  divided  into  any  integral  number  (n)  of  equal  parts, 
if  one  of  these  parts  is  contained  in  a  more  than  m,  times,  but 
less  than  m  +  1  times,  then 

«>l^but<!!L±l; 
b      n  r  n 

that  is,  the  value  of  ^'  lies  between  —  and  — ~ — 
6  n  n 

The  error,  therefore,  in   taking  either  of  these  values  for 

~  is  less  than  -.  But  by  increasing  n  indefinitely,  -  can  be 
b  n  n 

made   to  decrease  indefinitely,  and  to  become   less  than  any 

assigned  value,   however  small,  though    it   cannot   be   made 

absolutely  equal  to  zero. 

Hence,  the  ratio  of  two  incommensurable  quantities  cannot 

be  expressed  exactly  by  figures,  but  it  may  be  expressed  ap- 

p7-oa:tma<e/y*  within  any  assigned  measure  of  precision. 

255.  The  ratio  of  two  incommensurable  quantities  is  called 
an  inconirnciisurable  ratio  :  nud  is  n  fixed,  value  toward  which 
its  successive  approximate  values  constantly  tend. 


94  PLANE    GEOMETRY.  —  BOOK    II. 

256.  Theorem.  Two  incommensurahle  ratios  are  equal  ij, 
when  the  unit  of  measure  is  indefinitely  diminished,  their  ap- 
proximate values  constantly  remain  equal. 

Let  a:b  and  a' : 6'  be  two  incommensurable  ratios  whose  true 

values   lie   between   the   approximate    values  —  and  — ^tL_, 

n  n 

when  the  unit  of  measure  is  indefinitely  diminished.     Then 
they  cannot  differ  so  much  as    -• 

Now  the  difference  (if  any)  between  the  fixed  values  a  :  h 
and  a' :  b\  is  'a,  fixed  value.     Let  d  denote  this  difi'erence. 

Then  d<\. 

n 

But  if  d  has  any  value,  however  small,  -,  which  by  hypoth- 
esis can  be  indefinitely  diminished,  can  be  made  less  than  d. 

Therefore  d  cannot  have  any  value;  that  is,  d=0,  and 
there  is  no  difference  between  the  ratios  a :  b  and  a'  :b';  there- 
fore a:b  =  a' :  b\ 

The  Theory  of  Limits. 

257.  When  a  quantity  is  regarded  as  having  2i  fixed  value 
throughout  the  same  discussion,  it  is  called  a  constant;  but 
when  it  is  regarded,  under  the  conditions  imposed  upon  it,  as 
having  different  successive  values,  it  is  called  a  variable. 

When  it  can  be  shown  that  the  value  of  a  variable,  measured 
at  a  series  of  definite  intervals,  can  by  continuing  the  series 
be  made  to  differ  from  a  given  constant  by  less  than  any 
assigned  quantity,  however  small,  but  cannot  be  made  abso- 
lutely equal  to  the  constant,  that  constant  is  called  the  limit 
of  the  variable,  and  the  variable  is  said  to  approach  indefi- 
nitely to  its  limit. 

If  the  variable  is  increasing,  its  limit  is  called  a  superior 
limit ;  if  decreasing,  an  inferior  limit. 


THEORY    OF    LIMITS.  95 

Suppose  a  point  to  move  from  A  toward  B,  under  the  con- 
ditions that  the  first  ^  „  j^.  j^,  s 
second  it  shall  move  '  '  ■ 
one-half  the  distance  from  A  to  B,  that  is,  to  M\  the  next 
second,  one-half  the  remaining  distance,  that  is,  to  M^ ;  the 
next  second,  one-half  the  remaining  distance,  that  is,  to  M"  ; 
and  so  on  indefinitely. 

"  Then  it  is  evident  that  the  \noving  point  may  approach  as 
near  to  B  as  we  please,  but  will  nevm-  arrive  at  B.  Fr^r,  how- 
ever near  it  may  bef  to  B  at  an^  instant,  the  next  seco  id  it 
will  pass  ov«n  one-half  the  interyal  still  remaining ;  it  must, 
therefore,  approach  nearer  to  B,'  since  half  the  interval  still 
remaining  is  some  distance,  but  will  not  reach  B,  since  half 
the  interval  still  remaining  is  not  the  whole  distance. 

Hence,  the  distance  from  A  to  the  moving  point  is  an  in- 
creasing variable,  which  indefinitely  approaches  the  constant 
AB  as  its  limit;  and  the  distance  from  the  moving  point  to 
B  is  a  decreasing  variable,  which  indefinitely  approaches  the 
constant  zero  as  its  limit. 

If  the  length  of  AB  be  two  inches,  and  the  variable  be 
denoted  by  x,  and  the  difference  between  the  variable  and  its 
limit,  hj  v: 

after  one  second,  x=  1,  v.=  l 

after  two  seconds,        a;— 1-f-J,  v  =  ^ 

after  three  seconds,     ^'=l  +  -2-  +  i,  v  =  \ 

after  four  seconds,       ^"^l  +  j  +  i  +  i,  '^  —  i 
and  so  on  indefinitely. 

Now  the  sum  of  the  series  1  +  ^  +  ^  +  -g-,  etc.,  is  less  than 
2 ;  but  by  taking  a  great  number  of  terms,  the  sum  can  be 
made  to  differ  from  2  by  as  little  as  we  please.  Hence  2  is 
the  limit  of  the  sum  of  the  series,  when  the  number  of  the 
terms  is  increased  indefinitely ;  and  0  is  the  limit  of  the  dif- 
ference between  this  variable  sum  and  2. 


96 


PLANE   GEOMETRY.  —  BOOK    II. 


Consider  the  repetend  0.33333 ,  which  may  be  written 

A  +  TTU  +  Tinnr  +  TrlirTF  + 

However  great  the  number  of  terms  of  this  series  we  take, 
the  sum  of  these  terms  will  be  less  than  ^ ;  but  the  more 
terms  we  take  the  nearer  does  the  sum  approach  -J-.  Hence 
the  sum  of  the  series,  as  the  number  of  terms  is  increased. 
approaches  indefinitely  the  constant  -^^  as  a  limit.  ^ 

258.  In  the  right  triangle  A  CB,  if  the  vertex  A  approaches 
indefinitely  the  base  £C,  the  angle  £  a 
diminishes,  and  approaches  zero  indefi- 
nitely ;  if  the  vertex  A  moves  away  from 
the  base  indefinitely,  the  angle  B  increases 
and  approaches  a  right  angle  indefinitely ; 
but  B  cannot  become  zero  or  a  right  angle, 
so  long  as  ACB  is  a  triangle  ;  for  if  B  be-  ^' 
comes  zero,  the  triangle  becomes  the  straight  line  BC,  and  if 
B  becomes  a  right  angle,  the  triangle  becomes  two  parallel 
lines  ^Cand  AB  perpendicular  to  BC.  Hence  the  value  of 
B  must  lie  between  0°  and  90**  as  limits. 

259.  Again,  suppose  a  square  ABCD  inscribed  in  a  circle, 
and  E,  F,  H,  ^the  middle  points  of  the  arcs  subtended  by 
the  sides  of  the  square.  If  we  draw 
ihe  straight  lines  AE,  EB,  BE,  etc., 
we  shall  have  an  inscribed  polygon  of 
double   the   number   of  sides   of  the 


square. 

The  length  of  the  perimeter  of  this 
polygon,  represented  by  the  dotted 
lines,  is  greater  than  that  of  the 
square,  since  two  sides  replace  each 
side  of  the  square  and  form  with  it  a  triangle,  and  two  sides 
of  a  triangle  are  together  greater  than  the  third  side ;  but  less 
than  the  length  of  the  circumference,  for  it  '?-.  made  up  of 


THEORY    OF   LIMITS.  ^7 

straight  lines,  each  one  of  which  is  less  than  the  part  of  the 
circumference  between  its  extremities. 

By  continually  repeating  the  process  of  doubling  the  num- 
ber of  sides  of  each  resulting  inscribed  figure,  the  length  of 
the  perimeter  will  increase  with  the  increase  of  the  number 
of  sides ;  but  it  cannot  become  equal  to  the  length  of  the  cir- 
cumference, for  the  perimeter  will  continue  to  be  made  up  of 
straight  lines,  each  one  of  which  is  less  than  the  part  of  the 
circumference  between  its  extremities. 

The  length  of  the  circumference  is  therefore  the  limit  of  the 
Length  of  the  perimeter  as  the  number  of  sides  of  the  inscribed 
figure  is  indefinitely  increased. 

260.  Theorem.  //  two  variables  are  constantly  equal 
and  each  a])proiic7tes  a  lini'it,  their  limits  are  equal . 

V 


A 

X 

■c 

Let  AM  and  AN  be  two  variables  which  are  con- 
stantly equal  and  which  approach  indefinitely  AB 
and  AC  respectively  as  limits. 

To  prove  AB  =  AC. 

Proof.   If  possible,  suppose  A£  >  AC,  and  take  AD  =  AC. 

Then  the  variable  AM  ma,y  assume  values  between  ^Z>and 
ABj  while  the  variable  AI^  must  always  be  less  than  AB. 
But  this  is  contrary  to  the  hypothesis  that  the  variables  should 
continue  equal.  ' 

.'.  ^^  cannot  be  >  AC 

In  the  same  way  it  may  be  proved  that  ^C  cannot  be  >^  ^. 

.'.  AB  and  AC  are  two  values  neither  of  which  is  greater 
than  the  other 

Hence  A  H  -=  A  C. 


98  PLANE    GEOMETRY,  —  BOOK    II. 

Measure  of  Angles. 

Proposition  XVI.     Theorem. 

261.  In  tJve  same  circle,  or  equal  circles,  two  angles 
at  the  centre  have  the  same  ratio  as  their  intercepted 
arcs. 


Case  I.     When  the  arcs  are  commensurable. 
In  the  circles  whose  centres  are  C  and  D,  let  ACB  and 
EDF  be  the  angles,  AB  and  EF  the  intercepted  arcs. 

rp  Z  ACB       arc  AB 

10  prove  -^-z— =— -  = — — ,• 

^  Z  UBF     arc  EF     \ 

Proof.   Let  m  be  a  common  measure  oi  AB  and  FF. 

Suppose  m  to  be  contained  in  AB  seven  times, 
and  in  FF  four  times. 

Then  ^^^-l  (1) 

arc  ^i^      4  ^  ^ 

At  the  several  points  of  division  on  AB  and  FF  dY9i,w  radii. 
These   radii   will   divide   Z  A  CB    into   seven   parts,    and 
^  FDFinto  four  parts,  equal  each  to  each,  §  229 

(in  the  same  O,  or  equal  (D,  equal  arcs  subtend  eqiial  A  at  the  centre). 

.ZACB      7  (2) 


From  (1)  and  (2), 


ZFBF     4 

ZACB^&rcAB 
Z  FDF    arc  FF 


Ax.  1 


MEASURE    OF    ANGLES. 

Case  II.    When  the  arcs  are  incommensurable, 
p  P' 


99 


In  the  equal  circles  ABP  and  A'B'P'  let  the  angles 
ACB  and  A^C'B'  intercept  the  incommensurable  arcs 
AB  and  A!Bf. 

rj,  Z.  ACB        arc  AB 

To  prove  ____  =  ___, 

Proof.  Divide  AB  into  any  number  of  equal  parts,  and 
apply  one  of  these  parts  as  a  unit  of  measure  to  A^B^  as  many 
times  as  it  will  be  contained  in  A'B\ 

Since  AB  and  A^B^  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  from  -4'  to  some  point,  as  D,  leav- 
ing a  remainder  DB'  less  than  one  of  these  parts. 

Draw  C^D. 

Since  AB  and  A^D  are  commensurable, 

Z  ACB       arc  AB 


/.  A'C'D     arc  A'B 


Case  I. 


If  the  unit  of  measure  is  indefinitely  diminished,  these  ratios 
continue  equal,  and  approach  indefinitely  the  limiting  ratios 


Therefore 


Z  ACB        .   arc  AB 

and  — -• 

arc  A'B' 

arc  AB 


Z  A'C'B' 
ZACB 


§260 


Z  A'C'B'      arc  A'B' 

{If  two  vai'iables  are  constantly  equal,  and  each  approaches  a  limit,  theit 

limits  are  equal.) 


Q.  E.  O. 


100 


PLANE    GEOMETRY.  — BOOK    II. 


262.  The  circumference,  like  the  angular  magnitude  about 
a  point,  is  divided  into  360  equal  parts,  called  degrees.  The 
arc-degree  is  subdivided  into  60  equal  parts,  called  minutes ; 
and  the  minute  into  60  equal  parts,  called  seconds. 

Since  an  angle  at  the  centre  has  the  same  number  of  angle- 
degrees,  minutes,  and  seconds  as  the  intercepted  are  has  of  arc- 
degrees,  minutes,  and  seconds,  we  say :  A71  angle  at  the  centre 
is  measured  by  its  intercepted  arc;  meaning,  An  angle  at  the 
centre  is,  such  a  part  of  the  whole  angular  magnitude  about 
the  centre  as  its  intercepted  arc  is  of  the  whole  circumference. 

Proposition  XVII.     Theorem. 

263.  J^n  inscribed  angle  is  measured  iy  onechalf 
of  the  arc  intercepted  between  its  sides, 

B  B 


Case  I.    When  one  side  of  the  angle  is  Xi^diameter. 
In   the    circle  PAB  (Fig.  1),  let  'ilie  centre  C  be  in  - 
one  of  the  sides  of  the  inscribed  angle  B. 

To  prove         /.  B  is  measured  by  ^arc  PA. 
Proof.  Draw  CAl. 

Radius  CA  —  radius  CB. 

,'.ZB  =  ZA,  .  §154 

{being  opposite  equal  sides  of  the  A  CAB). 
But  ZFCA=ZB  +  ZA,  §145 

{the  exterior  Zofa  A  is  equal  to  the  sum  of  the  two  opposite  interior  A). 

.'.ZFCA  =  2ZB. 

But  Z  PC  A  is  measured  by  PA,  §  262 

{the  A  at  the  centre  is  measured  by  the  intercepted  arc). 

.'.  Z  B'm  measured  by  \  PA. 


MEASURE   OF   ANGLES.  101 

Case  II.    When  the  centre  is  within  the  angle. 
In   the  circle  BAE  (Fig.   2),  let    the    centre   C  fall 
within  the  angle  EBA. 
To  prove    Z  EBA  is  measured  hy  \  arc  EA. 
Proof.  Draw  the  diameter  EC  P. 

Z  PEA  is  measured  by  \  arc  PA,  Case  I. 

Z.  PEE  is  measured  by  -J  arc  PE,  Case  1. 

.-. Z  PEA  -r  Z  PEE  is  measured  by  \  (arc  PA  H-  arc  PE), 
or  Z  EEA  is  measured  by  \  arc  EA. 
Case  III.     When  the  centre  is  without  the  angle. 
In    the    circle   BFP   (Fig.    3),  let  the  centre  C  fall 
without  the  angle  ABF.  • 

To  prove     Z  AET  'is  measured  by  \  arc  AF. 
Proof.  ,    .  Draw  the' diameter  ECP. 

Z  PEE  is'measm-ed  by  ^  arc  PF,  Case  I. 

Z  ^EA  is  measui-ed  by  ^  arc  PA.  Case  I. 

.'.ZPEF-ZPEA  is-measured  by  |  (arc  PF-  &rc  PA), 
or  Z  AEF  '\8'm'ea,iiUTed  \>y  ^  arc  AF.  aE.D. 


Fig.  1.  Fia.  2.  Fig.  3. 

264.  Cor.  1.  An  angle  inscribed  m  a  semicircle  is  a  right 
angle.     For  it  is  measured  by  one-half  a  semi-circumference. 

265.  Cor.  2.  An  angle  inscribed  in  a  segment  greater  than  a 
semicircle  is  an  acute  angle.  For  it  is  measured  by  an  arc  less 
than  half  a  semi-circumference;  as.  Z  CAD.     Fig.  2. 

266.  Cor.  3.  An  angle  inscribed  in  a  segment  less  than  a 
semicircle  is  an  obtuse  angle.  For  it  is  measured  by  an  arc 
greater  than  half  a  semi-circumference  ;  as,  Z  CEE.     Fig.  2. 

267.  Cor.  4.  All  angles  inscribed  in  the  same  segment  are 
equal.     For  they  are  measured  by  half  the  same  arc.     Fig.  3. 


102  PLANE    GEOMETRY. BOOK    II. 


Peoposition  XVIIL     Theorem. 

268.  An  angle  formed  hy  two  chords  intersecting 
ivibhln  the  circumference,  is  measured  by  one-half 
the  sum  of  the  intercepted  arcs. 


Let  the  angle  AOC  be   kormed  by  the    chords  AB 
and  CD. 

To  prove    A  A  00  is  measured  hy  \  {A  C-\-  BD). 
Proof.  Draw  AD. 

ZCOA  =  ZD  +  AA,  §145 

{the  exterior  /.ofa/\is  equal  to  the  sum  of  the  two  opposite  interior  A). 

But  Z  D  k  measured  by  -J  arc  -4(7,  §  263 

ami  Z  A  is  measured  by  ^  arc  BD, 

{an  inscribed  Z  is  measured  by  ^  the  intercepted  arc). 

.-.  Z  CO  A  is  measured  by  i  {AC+  BD). 

Q.E,0. 


Ex.  83.    The  opposite  angles  of  an  inscribed  quadrilateral  are  sup- 
plements of  each  other. 
s5v    Ex.  84.    If  through  a  point  within  a  circle  two  perpendicular  chords 
''itre  drawn,  the  sum  of  the  opposite  arcs  which  they  intercept  is  equal  to 
a  semi-circumference. 

Ex.  85.  The  line  joining  the  centre  of  the  square  described  upon  the 
hypotenuse  of  a  rt.  A,  to  the  vertex  of  the  rt.  Z,  bisects  the  right  angle. 
Hint.  Describe  a  circle  upon  the  hypotenuse  as  diameter. 


MEASURE   OF   ANGLES. 


1U:J 


Froposition  XIX.     Theorem. 

269.  An  angle  formed  hy  a  tangent  arul  a  chord  is 
me<isitred  hy  one-half  the  intercepted  arc. 


Let  MAH  he   the  angle  formed  by  the  tangent  M<> 
and  chord  AH. 

To  prove    Z.  MA  His  rneasured  hy  \  arc  A  EH. 

Proof.      '  Draw  the  diameter  ACF. 

Z  Jf^i^  is  art.  Z,  §210 

{the  radius  drawn  to  a  tangent  at  the  'point  of  contact  is  JL  to  it). 

/L  MAF  being  a  rt.  Z,  is  naeasured  by  \  the  semi-circuni- 
ference  AEF. 

But  Z //yli^is  measured  by  J  arc //i^,  §263 

{an  inscribed  Z.  is  measured  by  h  the  intercepted  arc). 

.'.  Z  MAF-Z  HAFia  measured  by  ^{AFF-  HF)\ ^ 

or  Z  MAH  v^  measured  by  \  A  EH. 

Q.  e.  O. 

Ex.  80.  If  two  circles  touch  each  other  and  two  secants  are  drawn 
through  the  point  of  contact,  the  chords  joining  their  extremities  are 
parallel.     Hint.   Draw  the  common  tangent. 


104 


PLANE   GEOMETRY.  —  BOOK    II. 


Proposition   XX.     Theorem. 

270.  An  angle  formed  by  two  secants,  two  tangents, 
or  a  tangent  and  a  secant,  intersecting  without  the 
circumference,  is  measured  by  one-half  the  difference 
of  the  intercepted  arcs. 


Fig.  3. 


Cake  I.    Angle  formed  by  two  secafits. 

Let  the  angle  0  (Fig.  1)  be  formed  by  the  two  se- 
cants OA  and  OB. 

To  prove       Z.  0  is  measured  by  ^  (A^  —  JEC). 

Proof.  Draw  CB. 

/L  ACB^  A  0-Vl.B,  §  145 

{the  exterior  A  oj  a  IS  is  equal  to  the  sum  of  the  two  opposite  interior  A). 

By  taking  away  Z  B  from  both  sides, 

ZO^ZACB-ZB. 
^But 

and 


Z  A  CB  is  measured  by  \  AB, 


263 


Z  i?  is  measured  by  ^  CE, 
{an  inscribed  Z  is  measured  by  }  the  intercepted  arc). 


Z  0  is  measured  by  ^(AB—  CE), 


MEASURE   OF   ANGLES.  105 

§ASE  II.  Angle  formed  by  two  tangents. 

Let  the  angle  0  (Fig.  2)  be  formed  by  the  two  tan- 
gents OA  and  OB. 

To  prove  Z  0  is  measured  by  ^  {AMB  —  A8B). 

Proof.  ^  Draw  AB. 

/.  ABC=  ZO  +  Z  OAB,  %  145 

{the  exterior  Z  of  a  A  is  equal  to  the  sum  of  the  two  opposite  interior  A). 

By  taking  away  Z  OAB  from  both  sides, 

ZO  =  ZABC-ZOAB. 

But  •  Z  ABC  is  measured  by  ^  AMB,  §  269 

and  Z  OAB  is  measured  by  -J-  ASB, 

Can  Z  formed  hy  a  tangent  and  a  chord  is  measured  by  I  the  intercepted  arc). 

.-.  Z  0  is  measured  by  \  (AMB—  ASB). 

Case  III.    Angle  formed  by  a  tangent  and  a  secant. 

Let  the  angle  0  (Fig.  3)  be  formed  by  the  tangent 
OB  and  the  secant  OA. 

To  prove  Z  0  is  measured  by  -J-  (ADS  —  CES). 
Proof.  Draw  C8. 

ZACS=ZO  +  ZCSO/  §145 

(the  exterior  Zofa/\is  equal  to  the  sum  of  the  two  opposite  interior  A). 

By  taking  away  Z  C80  from  both  sides, 

Z0  =  ZAC8-ZC80. 

But  ZACSi^  measured  by  ^  ADS;  §  263 

(being  an  inscribed  Z), 

and  Z  CSO  is  measured  by  |  CES,  §  269 

(being  an  Z  formed  by  a  tangent  and  a  chord). 

.-.  Z  0  is  measured  hj  i(ADS—  CES). 

Q.E.O. 


106  PLANE   GEOMETRY. BOOK    II. 

Problems  of  Construction. 

Proposition   XXI.      Problem. 

271.   At  a  given  point  in  a  straight  line,  to  erect  a 
perpendicular  to  that  line. 


:^ 

* 

1         \ 

!                     1 

H      0      n 

C 

A- 

£; " -- - 

tM 

Fig.  I. 

Fig.  2. 

I.  Let  0  be  the  given  point  in  AC.  (Fig.  1). 
To  erect  a  JL  to  the  line  AC  at  the  point  0. 

Construction.  From  0  as  a  centre,  with  any  radius  OB, 
describe  an  arc  intersecting  ACin  two  points  .^Tand  J5. 

From  JTand  J5  as  centres,  with  equal  radii  greater  than 
OB,  describe  two  arcs  intersecting  at  B.     Join  OB. 

Then  the  line  OB  is  the  _L  required. 

Proof.    Since  0  and  B  are  two  points  at  equal  distances  from 

jETand  B,  they  determine  the  position  of  a  perpendicular  to 

the  line  ITB  at  its  middle  point  0.  §  123 

aE.F. 

II.  When  the  given  point  is  at  the  end  of  the  line. 
Let  B  be  the  given  point.  (Fig.  2). 

To  erect  a  JL  to  the  line  AB  at  B. 

Construction.  Take  any  point  C  without  AB ;  and  from  C 
as  a  centre,  with  the  distance  CB  as  a  radius,  describe  an  arc 
intersecting  AB  Sit  ^. 

Draw  BC,  and  prolong  it  to  meet  the  arc  again  at  D. 

Join  BD,  and  BB  is  the  X  required.  ' 

Proof.    The  Z  ^  is  inscribed  in  a  semicircle,  and  is  therefore 

a  right  angle.  §  264 

Hence  BB  is  -L  to  AB.  o.  e.  f. 


PROBLEMS. 


107 


Proposition   XXII.    '  Problem. 

272.   From  a  point  without  a  straight  line,  to  let 
fall  a  perpendicular  upon  that  line. 


KO 


H 


M 


Let  AB  be  a  given  straight  line,  and  C  a  given  point 
without  the  line. 

To  lei  fall  a  ±  to  the  line  AB  from  ike  •point  C. 

Construction.    From  C  as  a  centre,  with  a  radius  sufiiciently 
great,  describe  an  arc  cutting  AB  in  two  points,  ^and  K. 

From  ^and  K a.^  centres,  with  equal  radii  greater  than  \HK, 

describe  two  arcs  intersecting  at  0. 

Draw  CO, 

and  produce  it  to  meet  AB  at  M. 

CM'm  the  ±  required. 

Proof.  Since  Cand  0  are  two  points  equidistant  from  ^and 
K,  they  determine  a  _L  to  UK  at  its  middle  point.  §  123 

a  E.  F. 


Note.    Given  lines  of  the  figures  are  full  lines,   resulting  lines  are 
long-dotted,  and  auxiliary  lines  are  short-dotted. 


108  PLANE  GEOMETRY.  —  BOOK   II. 

Proposition  XXIII.    Problem. 
273.   To  bisect  a  given  straight  line. 


^I:i< 


4-^ 


# 


Let  AB  be  the  given  straight  line. 

To  bisect  the  line  AB. 

Construction.    From  A  and  B  as  centres,  with  equal  radii 
greater  than  J  AB,  describe  arcs  intersecting  at  Cand  E. 

Join  CE. 

Then  the  line  CE  bisects^^^. 

Proof.    C  and  E  are  two  points  equidistant  from  A  and  B. 
Hence  they  determine  a  X  to  the  middle  point  of  AB.    §  123 

a  E.  F. 


Ex.  87.   To  find  in  a  given  line  a  point  X  which  shall  be  equidis- 
tant from  two  given  points. 

Ex.  88.   To  find  a  point  X  which  shall  be  equidistant  from  two 
given  points  and  at  a  given  distance  from  a  third  given  point. 

Ex.  89.  To  find  a  point  X  which  shall  be  at  given  distances  from 
two  given  points. 

Ex.  90.  To  find  a  point  X  which  shall  be  equidistant  from  three 
given  points. 


PROBLEMS.  109 


Proposition  XXIV.   Probt.em. 
274.  To  bisect  a  $lven  arc. 


\  I  / 


Let  ACS  be  the  given  arc. 

To  bisect  the  arc  ACB. 

Oonstruction.  Draw  the  chord  AB, 

From  A  and  B  as  centres,  with  equal  radii  greater  than 
\  A  B,  describe  arcs  intersecting  at  I)  and  U, 

Draw  BR 

BE  bisects  the  arc  ACB. 

Proof.  Since  B  and  B  are  two  points  equidistamt  from  A 
and  B,  they  determine  a  JL  erected  at  the  middle  of  chord 
AB.  §  123 

And  a  X  erected  at  the  middle  of  a  chord  passes  through 
the  centre  of  the  O,  and  bisects  the  arc  of  the  chord.        §  234 

aE.  F. 

Ex.  91.  To  construct  a  circle  having  a  given  radius  and  passing 
through  two  given  points. 

Ex.  92.  To  construct  a  circle  having  its  centre  in  a  given  line  and 
passing  through  two  given  points. 


110 


PLANE    GEOMETHY. — BOOK    II. 


Proposition    XXV.     Problem. 
275.  To  bisect  a  given  angle. 


Let  AEB  he  the  given  angle. 

To  bisect  Z.  AEB. 

Construction.  From  ^as  a  centre,  with  any  radius,  as  EA, 
describe  an  arc  cutting  the  sides  of  the  A  E  at  A  and  B. 

From  A  and  B  as  centres,  with  equal  radii  greater  than 
one-half  the  distance  from  A  to  -6,  describe  two  arcs  inter- 
secting at  O. 

Jom  EC,  AC,  and  BC. 

^C  bisects  the  Z^. 

Proof.   In  the  A  ^i?C and  ^^(7 

AE=  BE,  and  AG=  BO,  Cons. 

and  EC=EC.  Iden. 

.-.  A  AEC==  A  BEC,  §  160 

{having  three  sides  equal  each  to  each). 

,:ZAEC=ZBEa 

Q.  E.  F. 


Ex.  93.   To  divide  a  right  angle  into  three  equal  parts. 
Ex.  94.    To  construct  an  equilateral  triangle,  having  given  one  side. 
Ex.  95.   To  find  a  point  Xwhich  shall  be  equidistant  from  two  given 
points  and  also  equidistant  from  two  given  intersecting  lines. 


PROBLEMS.  Ill 

Proposition    XXVI.     Problem. 

276.   At  a  given  point  in  a  given  straight  line,  to 
construct  an  angle  equal  to  a  ^iven  angle. 


Let  C  he  the  given  point  in  the  given  line  CM,  and 
A  the  given  angle. 

To  construct  an  A  at  Q  equal  to  the  Z  A. 

Oonstmctior.   From  -4  as  a  centre,  with  any  radius,  as  AE, 
describe  an  arc  cutting  the  sides  of  the  Z  ^  at  ^  and  F.  ■ 

From  Cas  a  centre,  with  a  radius  equal  to  AE, 

describe  an  arc  cutting  CJf  at  H, 

From  -ff  as  a  centre,  with  a  radius  equal  to  the  distance  EF, 

describe  an  arc  intersecting  the  arc  HG  at  m. 

Draw  Cm,  and  HCm  is  the  required  angle. 

Proof.  The  chords  ^i^and  Hm  are  equal.  Cons. 

.-.  arc  EF=  arc  Hm,  §  230 

{in  equal  <D  equal  chords  subtend  equal  arcs). 

.-.ZC^AA,  §229 

(in  equal  ©  equal  arcs  subtend  equal  A  at  the  centre).        q.  e.  f. 


V  Ex.  96.  In  a  triangle  ABQ  draw  DE  parallel  to  the  base  BO,  cut-  p  ^  i 
ting  the  sides  of  the  triangle  in  I)  and  E,  so  that  DE  shall  equal  3t^'^, 
DB±Ea 

Ex.  97.  If  an  interior  point  0  of  a  triangle  ^5Cis  joined  to  the  ver- 
tices B  and  C,  the  angle  BOO  is  greater  than  the  angle  BAC  of  the 
triande. 


112  PLANE   GEOMETRY. BOOK    II. 


Proposition    XXVII.     Problem. 

277,   Two  angles  of  a  triangle  bein^  given,  to  find 
the  third  anile. 


V 


n 


'1 V 

E ^ \ F 


Let  A  and  B  be  the  two  given  angles  of  a  triangle*. 

To  find  the  third  Z  of  the  A. 

Construction.   Take  any  straight  line,  as  EF,  and  at  any 
point,  as  H, 

construct  Z.  a  equal  to  Z  A^  §  276 

and  Z  b  equal  to  Z  B. 
Then  Z  c  is  the  Z  required. 

Proof.   Since  the  sum  of  the  three  ^  of  a  A  =  2  rt.  2^,  §  138 
and  the  sum  of  the  three  A  a,  6,  and  c,  =  2  rt.  ^;        §  92 
and  since  two  A  of  the  A  are  equal  to  the  A  a  and  5, 
the  third  Z  of  the  A  will  be  equal  to  the  Z  c.       ^     Ax.  3. 


Q.E.  F. 

Ex.  98.   In  a  triangle  ABC,  given  angles  A  and  B,  equal  respectively 
to  37°  13^  32^^  and  41°  W  56^''.    Find  the  value  of  angle  C. 


PROBLEMS. 


il3 


Proposition  XXVIII.  Problem. 

278.   Through  a  given  point,  to  draw  a  straight  line 
parallel  to  a  given  straight  line. 


H- 


C! 


Let  AB  be  the  given  line,  and  G  the  given  point. 
To  draw  through  the  point  C  a  line  parallel  to  the  line  AB. 
Construction.   Draw  DCE,  making  the  Z  EDB. 

At  the  point  O  construct  A  ECF=  Z  EDB.      §  27G 
Then  the  line  FCHi^  II  to  AB. 

Proof.  Z  ECF=  Z  EBB.  Cons. 

.-.  ^i^is  II  to^^,  §108 

{when  two  straight  lines,  lying  in  the  same  plane,  are  cut  by  a  third  straight 
line,  if  the  ext.-int.  A  are  equal,  the  lines  are  parallel). 

Q.E.  F. 


Ex.  99.  To  find  a  point  X  equidistant  from  two  given  points  and 

also  equidistant  from  two  given  parallel  lines. 

Ex.  100.    To  find  a  point  X  equidistant  from  two  given  intersecting 
lines  and  also  equidistant  from  two  given  parallels. 


114  PLANE    GEOMETRY. —BOOK    II. 


Proposition  XXIX.    Peoblem. 

279.    To  divide  a  given  straight  line  into    equal 
partfi. 

A  ^-- -. -jB 


€  "'•' 


•f) 


Let  AB  be  the  given  straight  line. 

To  divide  AB  into  equal  parts,  , 

Oonstruotion.        From  A  draw  the  line  AO. 

Take  any  convenient  length,  and  apply  it  to  AO  as  many 
times  as  the  line  AB  is  to  be  divided  into  parts, 

iFrom  the  last  point  thus  found  on  AO,  as  Q  draw  CB. 

Through  the  several  points  of  division  on  ^0  draw  lines 
II  to  CB,  and  these  lines  divide  AB  into  equal  parts. 

Proof.  Since  ACis  divided  into  equal  parts,  AB  is  also,  §  187 

[if  three  or  more  lis  intercept  equal  parts  on  any  transversal,  they  intercept 
equal  parts  on  every  transversal). 


Ex.  101.    To  divide  a  line   into  four  equal  parts  by  two  different 
methods. 

Ex.  102.   To  find  a  point  X  in  one  side  of  a  given  triangle  and  equi- 
distant from  the  other  two  sides. 

Ex.  103.   Through  a  given  point  to  draw  a  line  which  shall  make 
equal  angles  with  the  two  sides  of  a  given  angle. 


PROBLEMS,  f  -XV        ^ofi^         115 


Proposition    XXX.     Problem. 

280.   Two  sides  and  the  included  angle  of  a  trian- 
gle being  given,  to  construct  the  triangle. 

D 


h 


,4c 
/ !  \ 

-      /  ' 

A  I 

^z :c i...„ 

B 


Let  the  two  sides  of  the  triangle  be  b  and  c,  and  the 
included  angle  A. 

-To  construct  a  A  having  two  sides  equal  to  h  and  c  respec- 
tively, and  the  included  /.—Z.  A. 

Oonstruetion.   Take  AB  equal  to  the  side  c. 

At  A,  the  extremity  of  AB,  construct  an  angle  equal  to  the 
given  Z.A.  §  276 

On  AD  take  ^C  equal  to  b. 

Draw  CB. 

Then  A  A  CB  is  the  A  required. 


Ex.  104.    To  construct  an  angle  of  45°. 

Ex.  105.  To  find  a  point  X  wliich  shall  be  equidistant  from  two 
given  intersecting  lines  and  at  a  given  distance  from  a  given  point. 

Ex.  106.  To  draw  through  two  sides  of  a  triangle  a  line  ||  to  the 
third  side  so  that  the  part  intercepted  between  the  sides  shall  have  a 
given  length. 


116  PLANE   GEOMETRY.  — BOOK   II. 


Peoposition   XXXI.      Problem. 

281.   A  side  and  two  angles  of  a  triangle  being 
given,  to  rnnfitruct  the  triajigle. 


:::-.73 

; 

c 
Let  c  be  the  given  side,  A  and  B  the  given  angles. 

To  construct  the  triangle. 

Oonstruction.  Take  EC  equal  to  c. 

At  the  point  E  construct  the  ZC^^  equal  to  /.  A.     §  276 

At  the  point  C  construct  the  Z.  ECK  equal  to  Z  B. 

Let  the  sides  EJS"  smd  C^  intersect  at  O. 
Then  A  COE  is  the  A  required. 

Q.E.F. 

Remaek.   If  one  of  the  given  angles  is  opposite  to  the  given  side, 
find  the  third  angle  by  I  277,  and  proceed  as  above. 

Discussion.   The  problem  is  impossible  when  the  two  given 
angles  are  together  equal  to  or  greater  than  two  right  angles. 


Ex.  107.  To  construct  an  angle  of  150°. 

Ex.  108.  A  straight  railway  passes  two  miles  from  a  town.  A  place 
is  four  miles  from  the  town  and  one  mile  from  the  railway.  To  find  by 
construction  how  many  places  answer  this  description. 

Ex.  109.  If  in  a  circle  two  equal  chords  intersect,  the  segments  of  one 
chord  are  equal  to  the  segments  of  the  other,  each  to  each. 

Ex.  110.  AB  is  any  chord  and  J-Cis  tangent  to  a  circle  at  A,  CDE  a 
line  cutting  the  circumference  in  D  and  E  and  parallel  to  AB\  show 
that  the  triangles  ACD  and  EAB  are  mutually  equiangular. 


PROBLEMS.  117 


Proposition  XXXII.     Problem. 

282.   The  three  sides  of  a  triangle  heing  given,  to 
construct  the  triangle. 


/ 

/ 
/ 


4^ _ 1.B 

Let  the  three  sides  be  m,  n,  and  o. 

To  construct  the  triangle. 

Oonstruction.     '      Draw  AB  equal  to  o. 

From  ^  as  a  centre,  with  a  radius  equal  to  w,  describe  an 
arc ; 

and  from  ^  as  a  centre,  with  a  radius  equal  to  w,  describe 
an  arc  intersecting  the  former  arc  at  C. 

Draw  CA  and  CB. 

Then  A  CAB  k  the  A  required. 

Q.E.F 

Discussion.  The  problem  is  impossible  when  one  side  is  equal 
to  or  greater  than  the  sum  of  the  other  two. 

Ex.  111.  The  base,  the  altitude,  and  an  angle  at  the  base,  of  a  tri- 
angle being  given,  to  construct  the  triangle. 

Ex.  112.  Show  that  the  bisectors  of  the  angles  contained  by  the  oppo- 
site sides  (produced)  of  an  inscribed  quadrilateral  intersect  at  right  angles. 
^"  Ex.  113.  Given  two  perpendiculars,  AB  and  CD,  intersecting  in  0,  and 
a  straight  line  intersecting  these  perpendiculars  in  E  and  F;  to  construct 
a  square,  one  of  whose  angles  shall  coincide  with  one  of  the  right  angles 
at  O,  and  the  vertex  of  the  opposite  angle  of  the  square  shall  lie  in  EF. 
(Two  solutions.) 


118  PLANE    GEOMETRY. BOOK    II. 


Proposition  XXXlli.    Problem. 

283.    Two  sides  of  a  triangle  and  the  angle  opposite 
one  of  them,  being  given,  to  eonstruet  the  triangle. 


-6 


c 


Case  L  If  the  side  opposite  to  the  given  angle  is  less  than 
the  other  given  side. 

Let  h  be  greater  than  a,  and  A  the  given  angle. 

To  construct  the  triangle. 

Construction.    Constniet  Z  DAU  =  to  the  given  Z  A.  ^  27G 
On  AD  take  AB  =  b. 
From  i?  as  a  centre,  with  a  radius  equal  to  a, 
describe  an  arc  intersecting  the  side  ATJ  rI  (7  and  C. 
Draw  ^C  and  ^(7'. 

Then  both  the  A  ABC  and  ABC  yD 

fulfil  the  conditions,  and    hence  we  ,-■'' 

have    two    constructions.      This    is  y^ 

called  the  ambiguous  case.  l'^ 

Discussion.    If  the  side  a  is  equjii,  ___Z___Ji- 
to  the  _L  BH,  the  arc  described  frou; 
B  will  tquch  AE,  and  there  will  be 
but  one  construction,  the  right  tri- 
angle ABU.  B/' 

If  the  given  side  a  is  less  than  the  /"'  \a 

±  from  B,  the  arc  described  from  B  X     ^- 

will  not  intersect  or  touch  AE,  anri  — -^ 

hence  the  problem  is  irnpot;Ril)le. 


THE   CIRCLE.  119 

If  the  Z  ^  is  right  or  obtuge,  the  problem  is  impossible ;  for  the 
side  opposite  a  right  or  obtuse  angle  is  the  greatest  side.     §  159 

Case  II.  If  a  is  equal  to  b. 

If  the  Z  ^  is  acute,  and  a^=^h,  the  arc  described  from  B  as 
a  centre,  and  with  a  radius  equal  to  a,  will  - 
cut  the  line  AE  at  the  points  A   and  Jk  ji  / 

There  is   therefore   but   one  solution :    the  h/  '• « 

•  »  _. 

isosceles  A  ABC.  ^'--.. .,.--(}  ^ 

Discussion.   If  the  Z.  A\q  right  or  obtuse, 
the  problem  is  impossible:  for  equal  sides  of  a  A  liave  e(jUMl 
A  opposite  them,  and  a  A  cannot  have  two  right  A  or  two 
obtuse  A. 

Case  III.  If  a  is  greater  than  b. 

If  the  given  Z  ^  is  acute,  the  aic  described  from  B  will  cut 
the  line  ED  on  opposite  sides  of  ^,  at  (7  and  6''.  The  A  ABC 
answers  the  required  conditions,  but  the  ^  .y 

A  ABC  does  not,  for  it  does  not  contain      ^      a...   "  x^ 

the  acute  Z^.    There  is  then  only  one  e  V'  { ~^^ 

solution;  namely,  the  A  ^5C.  "^^-- --'' 

If  the  Z  A  is  right,  the  arc  described 
from  B   cuts   the   line  ED  on  opposite  z^x 

sides  of  A,  and  we  have  two  equal  right  "/   \h\ 


A  which  fulfil  the  required  conditions.  B  J^--.X- -c 

If  the  Z  ^  is  obtuse,  the  arc  described 
from  B  cuts  the  line   ED  on   opposite  \j5 

sides  of  A,  at  the  points  O  and  C.     The  a/  \a 


A  ABC  answers  the  required  conditions,   ^    \/  \  x^''  p 

but  the  A  ABC  does  not,  for  it  does         ^   "^ '' 

not  contain  the  obtuse  Z  A.     There  is  then  only  one  solu- 
tion ;  namely,  the  A  ABC. 

Q.  E.  F 


120  PLANE   GEOMETRY.  —  BOOK    II. 


Proposition  XXXIV.  Problem. 

284.   Two  sides  and  an  included  angle  of  a  paral- 
lelogram being  given,  to  construct  the  parallelogram. 


H/ 

/ 

/ 

/ 

b/ 

/ 

--,  / 

/ 

.K" 

/ 

/  \ 

/ 
/ 

A'—j 

'B 

Let  m  and  o  be  the  two  sides,  and  C  the  included 
angle. 

■To  construct  a  parallelogram. 

Construction.  Draw  AB  equal  to  o. 

At  A  construct  the  Z  A  equal  to  Z  C,  §  276 

and  take  ^^  equal  to  m. 

From  -S'as  a  centre,  with  a  radius  equal  to  o,  describe  an  arc. 

From  -5  as  a  centre,  with  a  radius  equal  to  m, 

describe  an  arc,  intersecting  the  former  arc  at  E. 

Draw  JEJir&nd  EB, 

The  quadrilateral  ABEH  is  the  O  required. 

Proof.  AB  =  HE,  Cons. 

AH=  BE.  Cons. 

.-.  the  figure  ABEHi^  a  O,  §  183 

{having  its  oppomXe  sides  equal). 

Q,  E.  F. 


PROBLEMS.  121 


Proposition  XXXV.     Problem. 

285.    To  circumscribe   a  circle  about   a  given  tri- 
angle, ^,_ 


Let  ABC  be  the  given  triangle. 

To  circumscribe  a  cii'cle  about  ABC. 

Ooiistruction.  Bisect  AB  and  BC.  §  273 

At  the  points  of  bisection  erect  Js.  §  271 

Since  BC  is  not  the  prolongation  of  AB,  these  Js  will  in- 
tersect at  some  point  0. 

From  0,  with  a  radius  equal,  to  OB,  describe  a  circle. 

O  vl^C  is  the  O  required. 

Proof.    The  point  0  is  equidistant  from  A  and  B, 

and  also  is  equidistant  from  B  and  C,  §  122 

[every  point  in  the  1.  erected  at  the  middle  of  a  straight  line  is  equidistant 
from  the  extremities  of  that  line). 

.'.  the  point  O  is  equidistant  from  A,  B,  and  C, 

and  a  O  described  from  0  as  a  centre,  with  a  radius  equal  to 
OB,  will  pass  through  the  vertices  A,  B,  and  C  aE.F. 

286.  Scholium.  The  same  construction  serves  to  describe  a 
circumference  which  shall  pass  through  the  three  points  not 
in  the  same  straight  iine  ;  also  to  find  the  centre  of  a  given 
circle  or  of  a  given  arc. 


122 


PLANE   GEOMETRY.  —  BOOK    II. 


Proposition    XXXVI.      Problem. 

287.    Through  a  given  point,  to  draw  a  tangent  to  a 
o^iven  circle. 


'--^M. 


rAE 


Case  I.    Wiioi  the  yiveu  point  ls  on  the  circle. 

Let  C  be  the  given  point  on  the  circle. 

To  draw  a  tangent  to  the  circle  at  C. 

Oonstruction.    From  the  centre  0  draw  the  radius  OC. 

Through  Cdraw  AMI.  to  OC.  §  271 

Then  ^Jfis  the  tangent  required. 
Proof.    A  straight  line  ±  to  a  radius  at  its  extremity  is  tan- 
gent to  the  circle.  §  239 

Case  II.  When  the  given  point  is  ivithout  the  circle. 
Let  O  be  the  centre  of  the  given  circle,  E  the  given 
point  without  the  circle. 

To  draw  a  tangent  to  the  given  circle  from  the  point  E. 
Oonstruction.  Join  OE. 

On  OE  as  a  diameter,  describe  a  circumference  intersecting 
the  given  circumference  at  the  points  Jf  and  H. 
Draw  OJf  and  EM. 
Then  EM  is  the  tangent  required. 

Proof.  Z  OME  is  a  right  angle,  §  264 

{being  inscribed  in  a  semicircle). 
.'.  EM  is  tangent  to  the  circle  at  M.  §  239 

In  like  manner,  we  may  prove  ITE  tangent  to  the  given  O. 

Q.  E.  P. 


PKOBLEMS. 


123 


Proposition   XXXVII.      Problem. 
To  inscribe  a  circle  in  a  given  triangle. 

B 


Let  ABC  be  the  given  triangle. 
To  iyiscribe  a  circle  in  the  A  ABC. 

Oonstniction.  Bisect  A  A  and  C.  §  275 

From  Uy  the  intersection  of  these  bisectors, 

draw  EII±  to  the  line  AC.  §  272 

From  U,  with  radius  UIT,  describe  the  O  KMH. 
The  O  I<:HM'\%  the  ©required. 
Proof.    Since  E  is  in  the  bisector  of  the  Z  A,  it  is  equidis- 
tant from  the  sides  AB  and  AC\  and  since  E  is  in  the  bisector 
of  the  Z  C,  it  is  equidistant  from  the  sides  ^Cand  EC,  §  162 
{every  point  in  the  bisector  of  an  Z  is  equidistant  from  the  sides  of  the  Z). 
.'.  a  O  described  from  E  ac  centre,  with  a  radius  equal  to 
E.E,  will  touch  the  sides  of  the  A  and  be  inscribed  in  it.. 


Q.  E.  F. 


289.  Scholium.  The  intersec- 
tions of  the  bisectors  of  exterior 
angles  of  a  triangle,  formed  by 
producing  the  sides  of  the  tri- 
angle, are  the  centres  of  three 
circles,  each  of  which  will  touch 
one  side  of  the  triangle,  and  the 
two  other  sides  produced.  These 
three  circles  are  called  escribed 
circles. 


124  PLANE    GEOMETRY. BOOK    II. 


Peoposition  XXXVIII.    Problem. 

290.    Upon  a  given  straight  line,  to  describe  a  seg- 
ment of  a  circle  which  shall  contain  a  given  angle. 


/     /  \ 

~W7    i^-.     V     ^/b 


/E 


Let  AB  be  the  given  line,  and  M  the  given  angle. 

To  describe  a  segment  upon  AB  which  shall  contain  Z.  M. 

Construction.  Construct  Z  ABE  equal  to  Z  if.  §  276 

Bisect  the  line  AB  by  the  _L  FO.  §  273 

From  the  point  B  draw  BO  ±  to  EB.  §  271 

From  O,  the  point  of  intersection  of  FO  and  BO,  as  a  cen- 
tre, with  a  radius  equal  to  OB,  describe  a  circumference. 
The  segment  AKB  is  the  segment  required. 

Proof.    The  point  0  is  equidistant  from  J.  and  ^,  §  122 

{every  point  in  a  ±  erected  at  the  middle  of  a  straight  line  is  equidistant 
from  the  extremities  of  that  line). 

.'.  the  circumference  will  pass  through  A. 

But  BE  is  ±  to  OB.  Cons. 

.-.  BE  is  tangent  to  the  O,  §  239 

(a  straight  line  J-  to  a  radius  at  its  extremity  is  tangent  to  the  O). 

.-.  Z  ABE  is  measured  by  2  arc  AB,  §  269 

(being  an  Z  formed  by  a  tangent  and  a  chord). 

An  Z  inscribed   in   the   segment  AKB   is   measured   by 

\AB.  §263 

.*.  segment  AKB  contains  Z  31.  Ax.  1 

aE.F. 


PROBLEMS.  125 


Proposition  XXXIX.    Problem. 


291.  To  find  the  ratio  of  two  commensurdble  straight 

/  H 
1— L-B 


Lines,  f/\ 

E   H 


K 

C' ^ r— ^D. 

Let  AB  and  CD  be  two  straight  lines.  '^ 

To  find  ike  ratio  of  AB  and  CD. 

Apply  CD  to  AB  SLS  many  times  as  possible. 

Suppose  twice,  with  a  remainder  UB. 

Then  apply  UB  to  CD  as  many  times  as  possible. 

Suppose  three  times,  with  a  remainder  FD.. 
Then  apply  FD  to  JEB^slB  many  times  as  possible. 

Suppose  once,  with  a  remainder  S'B. 
Then  apply  JIB  to  FD  as  many  times  as  possible. 

Suppose  once,  with  a  remainder  ICD. 

Then  apply  KD  to  JIB  as  many  times  as  possible. 

Suppose  J^D  is  contained  just  twice  in  JIB. 

The  measure  of  each  line,  referred  to  KD  as  a  unit,  will 
then  be  as  follows : 

IIB  =  2F:D; 

FD==  IIB-\-l^D=  SKD 
FB=  FD+JIB=  bKD 
CD  -=^EB-\-FD  =  l^KD 
AB=^2CD  -{-FB=4:1  KD 
.  AB_^\KD, 
"CD      18  XD' 

/.the  ratio  4^  =  — 

CD  18  Q.E.R 


V 


126  PLANE    GEOMETRY.  —  BOOK    II. 


Theorems. 

114.  The  shortest  line  and  the  longest  line  which  can  be  drawn  from 
a  given  point  to  a  given  circumference  pass  through  the  centre. 

115.  The  shortest  chord  that  can  be  drawn  through  a  given  point 
within  a  given  circle  is  ±  to  the  diameter  which  passes  through  the  point. 

116.  In  the  same  circle,  or  in  equal  circles,  if  two  arcs  are  each 
greater  than  a  semi-circumference,  the  greater  arc  subtends  the  less 
chord,  and  conversely. 

117.  If  ABC  is  an  inscribed  equilateral  triangle,  and  Pis  any  point 
in  the  arc  BC,  then  PA  -  PB  +  PC. 

Hint.   On  PA  take  PI/ equal  to  PP,  and  join  BM. 

118.  In  v/hat  kinds  of  parallelograms  can  a  circle  be  inscribed  ? 
Prove  your  answer. 

119.  The  radius  of  the  circle  inscribed  in  an  equilateral  triangle  is 
equal  to  one- third  of  the  altitude  of  the  triangle. 

120.  A  circle  can  be  circumscribed  about  a  rectangle. 

121.  A  circle  can  be  circumscribed  about  an  isosceles  trapezoid. 

122.  The  tangents  drawn  through  the  vertices  of  an  inscribed  rec- 
tangle enclose  a  rhombua. 

123.  The  diameter  of  the  circle  inscribed  in  a  rt.  A  is  equal  to  the 
difference  between  the  sum  of  the  legs  and  the  hypotenuse. 

124.  From  a  point  A  without  a  circle,  a  diameter  AOB  is  drawn, 
and  also  a  secant  ACD,  so  that  the  part  AC  without  the  circle  is  equal 
to  the  radius.     Prove  that  the  Z  I>AB  equals  one-third  the  Z  BOB. 

125.  All  chords  of  a  circle  which  touch  an  interior  concentric  circle 
are  equal,  and  are  bisected  at  the  points  of  contact. 

126.  If  two  circles  intersect,  ^nd  a  secant  is  drawn  through  each 
point  of  intersection,  the  chords  which  join  the  extremities  of  the  secants 
are  parallel.  Hint.  By  drawing  the  common  chord,  two  inscribed 
quadrilaterals  are  obtained. 

127.  If  an  equilateral  triangle  is  inscribed  in  a  circle,  the  distance  of 
each  side  from  the  centre  of  the  circle  is  equal  to  half  the  radius  of  the 
circle. 

128.  Through  one  of  the  points  of  intersection  of  two  circles  a 
diameter  of  each  circle  is  drawn.  Prove  that  the  straight  line  joining 
the  ends  of  the  diameters  passes  through  the  other  point  of  intersection. 


EXEKCISES.  127 

129.  A  circle  touches  two  sides  of  an  angle  BAC a,t  B,  C;  through  any 
point  D  in  the  arc  BC  a  tangent  is  drawn,  meeting  AB  at  J57and  AC 
at  F.  Prove  (i.)  that  the  perimeter  of  the  triangle  AEF  is  constant  for 
all  positions  of  D  in  BC;J^  that  the  angle  EOF  is  also  constant. 

Loci. 

130.  Find  the  locus  of  a  point  at  three  inches  from  a  given  point. 

131.  Find  the  locus  of  a  point  at  a  given  distance  from  a  given 
circumference. 

132.  Prove  that  the  locus  of  the  vertex  of  a  right  triangle,  having  a 
given  hypotenuse  as  base,  is  the  circumference  described  upon  the  given 
hypotenuse  as  diameter. 

133.  Prove  that  the  locus  of  the  vertex  of  a  triangle,  having  a  given 
base  and  a  given  angle  at  the  vertex,  is  the  arc  which  forms  with  the 
base  a  segment  capable  of  containing  the  given  angle. 

134.  Find  the  locus  of  the  middle  points  of  all  chords  of  a  given 
length  that  can  be  drawn  in  a  given  circle. 

135.  Find  the  locus  of  the  middle  ]K)int.s  of  all  chords  that  can  be 
•  Irawn  through  a  given  point  -4  in  a  given  circumference. 

136.  Find  the  locus  of  the  middle  points  of  all  secants  that  can  be 
lirawu  from  a  given  point  A  to  a  given  circumference^ 

137.  A  straight  line  moves  so  that  it  remains  parallel  to  a  given  line, 
and  touches  at  one  end  a  given  circumference.  Find  the  locus  of  the 
other  end. 

138.  A  straight  rod  moves  so  that  its  ends  constantly  touch  two 
fixed  rods  which  are  ±  to  each  other.   Find  the  locus  of  its  middle  point. 

139.  In  a  given  circle  let  A.OB  be  a  diameter,  OC  any  radius,  CD 
the  jierpendicular  from  Cto  AB.  Upon  OC  take  OM^CB.  Find  the 
locus  of  the  point  1/as  OC  turns  about  0. 

Construction  of  Polygons. 

To  construct  an  eqtftklteral  A,  having  given  : 

140.  The  perimeter.  141.   The  Vadius  of  the  circumscribed  circle. 
142.   The  altitude.  143.   The  radius  of  the  inscribed  circle. 

To  construct  an  isosceles  triangle,  having  given : 
144.    The  angle  at  the  vertex  and  the  ba«e. 


128  PLANE   GEOMETRY.  —  BOOK   II. 

145.  The  angle  at  the  vertex  and  the  altitude. 

146.  The  base  and  the  radius  of  the  circumscribed  circle. 

147.  The  base  and  the  radius  of  the  inscribed  circle. 

148.  The  perimeter  and  the  alti- 
tude. 

Hints.  Let  ABC  be  the  A  re- 
quired, and  EF  the  given  perimeter. 
The  altitude  CD  passes  through  the  ^^'< 

middle  of  EF,   and   the  ^  AEC, 
BFC&re  isosceles.  E  A    D    B  F 

To  construct  a  right  triangle,  having  given : 

149.  The  hypotenuse  and  one  leg. 

150.  The  hypotenuse  and  the  altitude  upon  the  hypotenuse. 

151.  One  leg  and  the  altitude  upon  the  hypotenuse  as  base. 

152.  The  median  and  the  altitude  drawn  from  the  vertex  of  the  rt.  Z. 

153.  The  radius  of  the  inscribed  circle  and  one  leg. 

154.  The  radius  of  the  inscribed  circle  and  an  acute  angle. 

155.  An  acute  angle  and  the  sum  of  the  legs. 

- 156.   An  acute  angle  and  the  difference  of  the  legs. 

To  construct  a  triangle,  having  given : 

157.  The  base,  the  altitude,  and  the  Z  at  the  vertex. 

158.  The  base,  the  corresponding  median,  and  the  Z  at  the  vertex. 

159.  The  perimeter  and  the  angles. 

^160.  One  side,  an  adjacent  Z,  and  the  sum  of  the  other  sides. 

161.  One  side,  an  adjacent  Z,  and  the  difference  of  the  other  sides. 

162.  The  sum  of  two  sides  and  the  angles. 

163.  One  side,  an  adjacent  Z,  and  radius  of  circumscribed  O. 

164.  The  angles  and  the  radius  of  the  circumscribed  O. 

165.  The  angles  and  the  radius  of  the  inscribed  O. 

166.  An  angle,  the  bisector,  and  the  altitude  drawn  from  the  vertex 

167.  Two  sides  and  the  median  corresponding  to  the  other  side. 

168.  The  three  medians. 

To  construct  a  square,  having  given : 

169.  The  diagonal.       <  170.   The  sum  of  the  diagonal  and  one  side. 


EXERCISES.  129 

To  construct  a  rectangle,  having  given : 

171.  One  side  and  the  Z  formed  by  the  diagonals. 

172.  The  perimeter  and  the  diagonal. 

173.  The  perimeter  and  the  Z  of  the  diagonals. 

174.  The  difference  of  the  two  adjacent  sides  and  the  Z  of  the 
diagonals. 

To  constmct  a  rhombus,  having  given : 

175.  The  two  diagonals. 

176.  One  side  and  the  radius  of  the  inscribed  circle. 

177.  One  angle  and  the  radius  of  the  inscribed  circle. 

178.  One  angle  and  one  of  the  diagonals. 

To  construct  a  rhomboid,  having  given: 

179.  One  side  and  the  two  diagonals. 

180.  The  diagonals  and  the  Z  formed  by  them. 

181.  One  side,  one  Z,  and  one  diagonal. 

182.  The  base,  the  altitude,  and  one  angle. 

To  construct  an  isosceles  trapezoid,  having  given: 

183.  The  bases  and  one  angle.        184.   The  bases  and  the  altitude. 

185.  The  bases  and  the  diagonal. 

186.  The  bases  and  the  radius  of  the  circumscribed  circle. 

To  construct  a  trapezoid,  having  given : 

187.  The  four  sides.      '  188.  The  two  bases  and  the  two  diagonals. 

189.  The  bases,  one  diagonal,  and  the  Z  formed  by  the  diagonals. 

Construction  of  Circles. 

Find  the  locus  of  the  centre  of  a  circle : 

190.  Which  has  a  given  radius  r  and  passes  through  a  given  point  P. 

191.  Which  has  a  given  radius  r  and  touches  a  given  straight  line  AB 

192.  Which  passes  through  two  given  points  Pand  Q. 

193.  Which  touches  a  given  straight  line  ^5  at  a  given  point  P. 

194.  Which  touches  each  of  two  given  parallels. 

195.  Which  touches  each  of  two  given  intersecting  lines. 


130  PLANE   GEOMETEY.  —  BOOK   II. 

To  construct  a  circle  which  has  the  radius  r  and  which  also : 

196.  Touches  each  of  two  intersecting  lines  AB  and  CD. 

197.  Touches  a  given  line  AB  and  a  given  circle  K 

198.  Passes  through  a  given  point  P  and  touches  a  given  line  AB. 

199.  Passes  through  a  given  point  P  and  touches  a  given  circle  K. 

To  construct  a  circle  which  shall : 

200.  Touch  two  given  parallels  and  pass  through  a  given  point  P. 

201.  Touch  three  given  lines  two  of  which  are  parallel. 

202.  Touch  a  given  line  AB  at  P  and  pass  through  a  given  point  Q. 

203.  Touch  a  given  circle  at  Pand  pass  through  a  given  point  Q. 

204.  Touch  two  given  lines  and  touch  one  of  thorn  at  a  given  point  P 

205.  Touch  a  given  line  and  touch  a  given  circle  at  a  point  P. 

206.  Touch  a  given  line  AB  at  P  and  also  touch  a  given  circle. 

207.  To  inscribe  a  circle  in  a  given  sector. 

208.  To  construct  within  a  given  circle  three  equal  circles,  so  that 
each  shall  touch  the  other  two  and  also  the  given  circle. 

209.  To  describe  circles  about  the  vertices  of  a  given  triangle  as 
centres,  so  that  each  shall  touch  the  two  others. 

Construction  of  Straight  Lines. 

210.  To  draw  a  common  tangent  to  two  given  circles. 

211.  To  bisect  the  angle  formed  by  two  lines,  without  producing  the 
lines  to  their  point  of  intersection. 

212.  To  draw  a  line  through  a  given  point,  so  that  it  shall  form  with 
the  sides  of  a  given  angle  an  isosceles  triangle. 

213.  Given  a  point  P  between  the  sides  of  an  angle  BAC.     To  draw 
through  P  a  line  terminated  by  the  sides  of  the  angle  and  bisected  at  P. 

214.  Given  two  points  P,  Q,  and  a  line  AB ;  to  draw  lines  from  P 
and  Q  which  shall  meet  on  AB  and  make  equal  angles  with  AB. 

Hint.    Make  use  of  the  point  which  forms  with  P  a  pair  of  points 
svmmetrical  with  respect  to  AB. 

215.  To  find  the  shortest  path  from  Pto  Q  which  shall  touch  a,VmeAB. 

216.  To  draw  a  tangent  to  a  given  circle,  so  that  it  shall  be  parallel 
to  a  given  straight  line. 


Jl^^  -U  P  IV  9  ^« 


BOOK  III. 

PROPORTIONAL.    LINES    AND    SIMILAR 
POLYGONS. 


The  Theory  of  Proportion. 

292.  A  proportion  is  an  expression  of  equality  between  two 
equal  ratios. 

A  proportion  may  be  expressed  in  any  one  of  the  follow- 
ing forms : 

7=  —  ;     a:6  =  c:a;     a\h  '.  '.  c  :  d\ 
b      a 

and  is  read,  "  the  ratio  of  a  to  5  equals  the  ratio  of  e  to  d." 

293.  The  terms  of  a  proportion  are  the  four  quantities  com- 
pared ;  the  first  and  third  terms  are  called  the  antecedents,  the 
second  e^nd  fourth  terms,  the  consequents;  the  first  and  fourth 
terms  are  called  the  extremes,  the  second  and  third  terms,  the 
means. 

294.  In  the  proportion  a:h  =  c\d,  d  is  a  fourth  propor- 
tional to  a,  h,  and  c. 

In  the  proportion  a:h  =  h:c,  c  is  a  third  proportional  to 
a  and  h. 

In  the  proportion  a:b  —  h\c,  h  is  a  mean  proportional 
between  a  and  c. 


132  PLANE   GEOMETRY.  —  BOOK    III. 

Proposition  I. 

295.  In  every  proportion  the  product  of  the  extremes 
is  eqii^al  to  the  product  of  the  means. 

Let  a:  b  =  c-  d. 

To  prove  ad^bc. 

Now  ^  =  ^. 

b     d 

whence,  by  multiplying  both  sides  by  bdy 

ad  =  be.  Q,  E,  o^ 

Proposition  II. 

296.  A  mean  proportional  between  two  quantities 
Is  equal  to  the  square  root  of  their  product. 

In  the  proportion  (t:b  —b  :c, 

h"  =  ac,  §  296 

{tilt  product  of  the  extremes  is  equal  to  the  product  of  the  means). 

Whence,  extracting  the  square  root, 

b  =  -Vac.  Q.  E.  D. 

Proposition  III. 

297.  //  the  product  of  two  quantities  is  equal  to  the 

product  of  two  others,  either  two  may  be  made  the 

extremes  of  a  pj^oportion  in  which  the  other  two  are 

made  the  means.  ^_ 

Let.  ad^  he. 

To  prove  a  :  b  =  c  :  d. 

Divide  both  members  of  the  given  equation  by  bd. 

Then'  ^=|, 

b      d 

©r,  a:b  =  c:  d.  a  e.  o. 


THEORY    OF   PROPORTION.  133 


Proposition  IV. 

298.  If  four  quantities  of  the  same  hind  are  in  pro- 
portion, they  will  he  in  proportion  by  alternation ; 
that  is,  the  first  term  will  he  to  the  third  as  the  sec- 
ond to  the  fourth.  * 

Let  a:h  =  c:d. 


Now  "-     " 


To  prove  a:  c  =  b  :d. 

a  _c 
h~  d 

Multiply  each  member  of  the  equation  by  -. 

c 

Then  ?  =  1 

c      d 

.  a:c  =  b:d. 


aE.D. 


Proposition  V. 

299.  If  four  quantities  are  in  .proportion,  they  will 
he  in  proportion  hy  inversion ;  that  is,  the  second  term 
will  he  to  the  first  as  th\fourth  to  the  third. 

Let  a:  b  =  c:  d. 

To  prove  b:a  =  d:c. 

Now  be  =  ad.  §  295 

Divide  each  member  of  the  equation  by  ac. 

Then  .         *=^, 

a     c  T 

or,  h  :  a  —  d :  c. 

Q.E.O. 


134  PLANE    GEOMETRY.  —  BOOK    III. 

Proposition  VI. 

300.   If  four  quantities  are  in  proportion,  they  will 
he  in  proportion  by  composition ;  that  is,  the  sum  of 
the  first  two  terms  will  he  to  the  second  term  as  the 
sum  of  the  last  two  terms  to  the  fourth  term. 
Let  a:h  =  G:d. 

To  jprove  a  -\-h  \h  ^=  c  -\-  d  \  d. 

Now  f  =  2 

b      d 

Add  1  to  each  member  of  the  equation. 

Then  l+l^l+l; 

b  d 

a-\-h  __e-\-d 


Q.E.  D. 


that  is,  —^  ,    . 

b  d 

or,  a  -\-  h  \h  =  c  -\-  d  \  d. 

In  like  manner,      a-^h  \a=^c-\- d\c. 

^  Proposition  VII. 

301.  If  foy^  quantities  are  in  proportion,  they  will 
he  in  proportion  by  division ;  that  is,  the  difference 
of  the  first  two  terms  will  be  to  the  second  term  as 
the  difference  of  the  last  two  terms  to  the  fourth 


term. 

Let  a:  b  =  c:d. 

To  prove 

a~b  :b  =  c~  d:  d. 

Now 

a  _c 
■  b~d 

Subtract  1  from  each  member  of  the  eq 

Then 

b             d 

that  is, 

a—  b      c  —  d 
b             d    '  ' 

or, 

a~b  :b  =  c  —  d :  d. 

In  like  manner, 

a  —  h\a  =  c~d\c. 

Q.E.DI 


THEORY    OF    PROPORTION.  135 

Proposition  VIII. 

302.   In  any  propoiiiion  the  terms  are  in  proportion 
by  composition  and  division ;  that  is,  the  sum  of  the 
first  two  terms  is  to  their  difference  as  the  sum  of 
the  last  two  term^s  to  their  difference. 
Let  a:b  =  c:d. 

a-\-b  _c-\-d' 

a  c 

a  —  b      c  —  d 


Then,  by  §  300, 
And,  by  §301, 
By  division, 


a  c 

ar-\-b  _c-\-d 
a  ~b      c  —  d 


or,  a-\-b:a  —  b—c-\-d\c  —  d. 


aE.  D. 


Proposition  IX. 

^  303.  In  a  series  of  equal  ratios,  the  sum  of  the  an- 
tecedents is  to  the  sum  of  the  consequents  as  any 
antecedent  is  to  its  consequent. 

Let  a:h  =  c:(l=e  :f=  g :  ft. 
To  prove     a-^c-\-e-{-g  :b-\-d-{-f-\- h  =  a:b. 
Denote  each  ratio  by  r. 

Whence,      a  =  br,     c  =  dr,     e  =fr,    g  =  hr. 
Add  these  equations. 

Then  a-^c-\-e^  g  =  (b-\-d-\-f-\-h)r. 

Divide  by  {b  +  d-\-f-\-  h). 


Then  pLp:A^^r  =  ^, 

b+d+f+h  b 

or,  a-\-c-\-e-}-g:b-\-d-{-f-\-h  =  a:b. 


Q.  E.  D 


136  PLANE    GEOMETRY.  —  BOOK    III. 

Proposition  X. 

304.   The  products  of  the  corresponding  terms  of 
two  or  more  proportions  are  in  proportion. 
Let  a:b  =  c:d,    e:f=g:h,    k:l  =  m:n. 
To  prove  aeh  :  bfl  =  cgm  :  dhn. 

Now  ?  =  ^,    1  =  2,    h=^. 

0      d   f      hi      n 

Whence,  by  multiplication, 

aeh  _  cgm 

bfl      dhn 

or,  ■  aeh :  bfl  =  cgm :  dhn. 


Q.E.  D. 


Proposition  XI. 


305.  Like  powers,  or  like  roots,  of  the  tenrbs  of  a 
proportion  are  in  proportion. 

Let  aih  —  cd. 

To  prove  a'\b''^c^:  d'\ 

-     L       11 
and  a»  :  bri  =  c^  :  c?«. 

Now  ?  =  1. 

b      d 

By  raising  to  the  nth  power, 

^  =  ~;  or  a^:b^  =  'c^:di 
6**     d^ 

By  extracting  the  nth  root, 

-y  =  — -]  or,   a*^  :  0^  =  c^  :  an. 

m  a  E.  D. 

306i  Equimultiples  of  two  quantities  are  the  products  ob- 
tained by  multiplying  each  of  them  by  the  same  number. 
Thus,  ma  and  mb  are  equimultiples  of  a  and  b. 


THEORY   OF   PROPORTION.  137 

Proposition  XII. 

307.  Equimultiples  of  two  quantities  are  in  the 
same  ratio  as  the  quantities  themselves. 

Let  a  and  b  be  any  two  quantities. 

To  prove  ma  :  7nb  =a:b. 

m 

Now  ?  =  ?. 

0         0 

Multiply  both  terms  of  first  fraction  by  m. 

Then  V^^=% 

mb      b 

or,  ma  :  mb  ^=  a  -.b. 

Q.  E.  D. 

308.  Scholium.  In  the  treatment  of  proportion  it  is  as- 
sumed that  fractions  may  be  found  which  will  represent  the 
ratios.  It  is  evident  that  the  ratio  of  two  quantities  may  be 
represented  by  a  fraction  when  the  two  quantities  compared 
can  be  expressed  in  integers  in  terms  of  a  common  unit.  But 
when  there  is  no  unit  in  terms  of  which  both  quantities  can  be 
expressed  in  integers,  it  is  possible  to  find  a  fraction  that  will 
represent  the  ratio  to  any  required  degree  of  accuracy.  (See 
§§  251-256.) 

Hence,  in  speaking  of  the  product  of  two  quantities,  as  for 
instance,  the  product  of  two  lines,  we  mean  simply  the  product 
of  the  numbers  which  represent  them  when  referred  to  a  com- 
mon unit. 

An  interpretation  of  this  kind  must  be  given  to  the  product 
of  any  two  quantities  throughout  the  Geometry. 


138 


PLANE    GEOMETRY.  —  BOOK    III. 


Proportional  Lines. 

Proposition  I.     Theorem. 

309.  If  a  line  is  drawn  through  two  sides  of  a  tri- 
angle parallel  to  the  third  side,  it  divides  those  sides 
proportio  nally. 


E/ \V  E/  \F 

B  Fig.  1.  C  Fig.  2. 

In  the  triangle  ABC  let  EF  be  drawn  parallel  to  BG. 

rj.  EB     FC 

To  prove  aE=  AF 

Case  I.   When  AE  and  EB  (Fig.  1)  are  commensurable. 

Find  a  common  measure  of  AE  and  EB,  as  BM. 

Suppose  BM  to  be  contained  in  BE  three  times, 

and  in  AE  four  times. 

E=i  (^) 

At  the  several  points  of  division  on  BE  and  AE  draw 
straight  lines  II  to  BC. 

These  lines  will  divide  ^Cinto  seven  equal  parts,  of  which 
i^Cwill  contain  three,  and  ^i^will  contain  four,  §  187 

{if  parallels  intercept  equal  parts  on  any  transversal,  they  intercept  equal 
parts  on  every  transversal). 

'     .  !£  =  ?.  (2) 

"  AF     4:  ^  ^ 

Compare  (1)  and  (2), 

EB^IV^  j^^^ 

AE     AF 


PROPORTIONAL    LINES.  139 

Case  II.  When  AE  and  EB  (Fig.  2)  are  incommensurable. 

Divide  AE  into  any  number  of  equal  parts,  and  apply  one 
of  these  parts  as  a  unit  of  measure  to  EB  as  many  times  as  it 
will  be  contained  in  EB. 

Since  ^^  and  EB  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  from  ^  to  a  point  K^  leaving  a 
remainder  KB  less  than  t  he  unit  of  measure. 
Draw  KH II  to  BQ. 

Then  ff=Tf-  ^^««  I' 

AE      AF 

Suppose  the  unit  of  measure  indefinitely  diminished,  the 

ratios  -^-—  and  — -  continue    equal ;   and    approach   indefi- 

EB  FG 

nitely  the  limiting  ratios  -— —  and  — --,  respectively. 
AE  AF 

Therefore  ^  =  ^'  §260 

ae.  D. 

310.  Cor.  1.  One  side  of  a  tiiangle  is  to  either  part  cut  off 
by  a  straight  line  parallel  to  the  base  as  the  other  side  is  to  the 
corresponding  part. 

For  EB:AE=FC:  AF,  by  the  theorem. 

.-.  EB:\-  AE\  AE=  FC+  AF:  AF,  §  300 

or  AB:AE=AC:AF 

311.  Cor.  2.  If  two  lines  are  cut  by  any  number  of  parallels, 
the  corresponding  intercepts  are  proportional. 

Let  the  lines  be  AB  and  CD. 

Draw  AN  II  to  CD,  cutting  the  lis  at  L,  M, 
and  N.     Then 

AL=CO,  LM=^GK,  MN=KD.    §180 

By  the  theorem,  B     n       d 

AS'.  AM=  AF:  AL  =  FH:  LM=^  HB  \  MN. 
That  is,  AF:CG=  FH:  QK^  HB :  KD. 

If  the  two  lines  AB  and  CD  were  parallel,  the  correspond- 
ing intercepts  would  be  equal,  and  the  above  proportion  be  true. 


A 

c 

FJ\L 

V 

H     \m 

V 

140  PLANE    GEOMETRY.  —  BOOK    III. 


Proposition  II.     Theorem. 

312.   If  a  straight  line  divide  two  sides  of  a  tri- 
angle proportionally,  it  is  parallel  to  the  third  side. 


In  the  triangle  ABC  let  EF  be  drawn  so  that 

AB^AC 
AE     AF 

To  prove  EFWtoBC. 

Proof.  From  E  draw  EH  II  to  BC. 

Then  AB:AE=^AC:AII,  '  §310 

\ 
{one  side  of  a  A  is  to  either  part  cut  off  by  a  line  II  to  the  base,  as  the  other 
side  is  to  the  corresponding  part). 

But  AB  :  AE=AC:  AF.  Hyp. 

•The  last  two  proportions  have  the  first  three  terms  equal, 
each  to  each ;  therefore  the  fourth  terms  are  equal ;  that  is, 

AF=  AH. 

.'.  EFsiTid  -£'.5"  coincide. 

But     .  ^^is  II  to  BC.  Cons. 

.'.  EF,  which  coincides  with  EH,  is  (j  to  BC. 

a  ED 


PROPORTIONAL    LINES.  141 


Proposition  III.    Theorem. 

313.  The  bisector  of  an  angle  of  a  triangle  divides 
the  opposite  side  into  segments  proportional  to  the 
other  two  sides. 


-.0 


B 

Let  CM  bisect  the  angle  C  of  the  triangle  GAB. 
To  prove  MA  :  MB  =  CA  :  CB. 

Proof.   Draw  AE  II  to  CMio  meet  -B(?  produced  at  E. 


Since  CM  is  II  to  AE  of  the  A  BAE,  we 

have 

§309 

MA.MB^CE'.CB. 

(1) 

Since  CM\^  II  to  AE, 

ZACM-=ZCAE, 

§104 

(being  alt.  int.  A  of  II  lines) ; 

id                            ZBCM=ZCEA, 

§106 

{being  ext.-int.  A  o/ll  lines). 

But                     theZAC3f=ZBCM. 

Hyp. 

.-.the  2::  CAE  =  A  CEA. 

Ax.  1 

.:  CE=  CA, 

§156 

[if  two  A  of  a  A  are  equal,  the  opposite  sides 

are  equal). 

Putting  CA  for  CE  in  (1),  we  have 

MA:MB=CA:C^. 

aE.o 


142  PLANE    GEOMETRY.  —  BOOK    III. 


Proposition  IV.     Theorem. 

314.  The  bisector  of  an  exterior  angle  of  a  triangle 
meets  the  opposite  side  produced  at  a  point  the  dis- 
tances of  which  from  the  extremities  of  this  side  are 
proportional  to  the  other  two  sides. 


Let  CM'  bisect  the   exterior  angle  ACE  of  the  tri- 
angle CAB,  and  meet  BA  produced  at  M'. 

To  prove  M'A  :  M^B  ^  CA  :  CB. 

Proof.        Draw  AF  II  to  CM"  to  meet  BC  at  R 
Since  AF\^  II  to  CM'  of  the  A  BCM\  we  have  §  309 

M'A'.M'B=CF-CB.  (1) 

Since  ^i^  is  II  to  CM\ 

t\iQAM'CE  =  A.AFC,  §106 

{being  ext.-mt.  AqfW  lines)  ; 

and  the  Z  M'CA  =  Z  CAF,  §  104 

{being  alt.-mt  AofW  lines). 

Since  OM'  bisects  the  Z  FCA, 

ZM'CF=ZM'CA. 

.'.  the  ZAFC  =  Z  CAF.  Ax.  1 

.-.  CA  =  CF,  §  156 

{if  two  A  of  a  A  are  equal,  the  opposite  sides  are  equal). 
Putting  CA  for  Ci^in  (1),  we  have 

M'A:M'B=CA:CB. 

Q.  E.  D, 


PROPORTIONAL    LINES.  143 

315.  Scholium.  If  a  given  line  AB  is  divided  at  M,  a 
point  between  the  extremities  A  and  B,  it  is  said  to  be 
divided  internally  into  the  segments  MA  and  MB ;  and  if  it 
is  divided  at  M\  a  point  in  the  prolongation  of  AB,  it  is  said 
to  be  divided  externally  into  the  segments  M^ A  and  M^B. 

JI//C , B 

A  M 

In  either  case  the  segments  are  the  distances  from  the  point 
of  division  to  the  extremities  of  the  line.  If  the  line  is  divided 
internally,  the  sum  of  the  segments  is  equal  to  the  line ;  and 
if  the  line  is  divided  externally,  the  difference  of  the  segments 
is  equal  to  the  line. 

Suppose  it  is  required  to  divide  the  given  line  AB  inter- 
nally and  externally  in  the  same  ratio ;  as,  for  example,  the 

ratio  of  the  two  numbers  3  and  5. 
\^  1 


M' 


m'  '  '  '  h 


"We  divide  AB  into  5  +  3,  or  8,  equal  parts,  and  take  3 
parts  from  A  ;  we  then  have  the  point  M,  such  that 

MA:MB  =  S:6.  (1) 

Secondly,  we  divide  AB  into  two  equal  parts,  and  lay  off 
on  the  prolongation  of  AB,  to  the  left  of  A,  three  of  these 
equal  parts ;  we  then  have  the  point  M\  such  that 

M'A:M'B  =  S:b.  (2) 

Pomparing  (1)  and  (2), 

MA:MB=^  MA  :  M'B. 

316.  If  a  given  straight  line  is  divided  internally  and 
externally  into  segments  having  the  same  ratio,  the  line  is 
said  to  be  divided  harmonically. 


144  PLANE    GEOMETRY. — BOOK    III. 

317.  Cor.  1.  The  bisectors  of  an  interior  angle  and  an  exte- 
rior angle  at  one  vertex  of  a  triangle  p 

divide  the  opposite  side  harmoni- 
cally. For,  by  §§  313  and  314,  each 
bisector  divides  the  opposite  side 
into  segments  proportional  to  the 
other  two  sides  of  the  triangle. 

318.  Cor.  2.  If  the  points  M  and  M'  divide  the  line  AB 
harmonically,  the  points  A  and  B  divide  the  line  MM^  har- 
monically. 

For,  if  MA:  MB  =  M'A  :  M'B, 

by  alternation,      MA  :  MA  =  MB  :  M^B.  §'  298 

That  is,  the  ratio  of  the  distances  of  A  from  M  and  M^  is 
equal  to  the  ratio  of  the  distances  of  B  from  M  and  M\ 

The  four  points  A,  B,  M,  and  M  are  called  harmonic 
points,  and  the  two  pairs,  A,  B,  and  M,  M,  are  called  con- 
jugate harmonic  points. 

Similar  Polygons. 

319.  Similar  polygons  are  polygons  that  have  their  homol- 
ogous angles  equal,  and  their  homologous  sides  proportional. 


E  1)  -£" 

Thus,  if  the  polygons  ABODE  and  A*B^C^D^E^  are  similar 

the  A  A,  B,  C,  etc.,  are  equal  to  A  A',  B\  C\  etc. 
,  AB       BQ        CD      , 

320.    In  two  similar  polygons,  the  ratio  of  any  two  iiomol- 
ogous  sides  is  called  the  ratio  of  similitude  of  the  polygons. 


SIMILAR   TRIANGLES.  145 

Similar  Triangles-. 
Proposition  V.     Theorem. 

321.  Two  mutually  equiangular  triangles  are  sim- 
ilar. 

A 

A  ^' 


In  the  triangles  ABC  and  A'B'C  let  angles  A,  B,  C  be 
equal  to  angles  A',  B',  C  respectively. 

To  prove  A.  ABO  and  A}B^C^  similar. 

Proof.         Apply  the  A  A'B'C  to  the  A  ABC, 

so  that  Z  A'  shall  coincide  with  Z  A. 

Then  the  A  A'B'C^  will  take  the  position  of  A  AEIT. 

Now  Z  AEir(s&me  as  Z.B')  =  Z  B. 

.-.  ^^is  II  to  BC,  §108 

{when  two  straight  lines,  lying  in  the  same  plane,  are  cut  by  a  third  straight 
line,  if  the  ext.-int.  A  are  equal  the  lines  are  parallel). 

:.  AB  :  AE=  AC:  AH,  §  310 

oi-  AB'.A'B'^AC'.A'C. 

In  like  manner,  by  applying  A  A'B'C*  to  A  ABC,  so  that 
Z  J5'  shall  coincide  with  Z  B,  we  may  prove  that 
AB-.A'B'^BC.B'C. 
Therefore  the  two  A  are  similar.  §  319 

Q.E.D. 

322.  Cor.  1.   Two  triangles  are  similar  if  two  angles  of  the 
one  are  equal  respectively  to  two  angles  of  the  other. 

323.  Cor.  2.    Two  right  triangles  are  similar  if  an  acute 
angle  of  the  07ie  is  equal  to  an  acute  angle  of  the  other. 


146 


PLANE    GEOMETRY. 


BOOK    III. 


Proposition  VI.     Theorem. 

324.   If  two  triangles  have  their  sides  respectively 
proportional,  they  are  similar. 


In  the  triangles  ABC  and  A'B'G'  let 
AB  ^  AG  ^  EG 
A'B'     A'Gf     B'G'' 

To  prove  A  ABO  and  A^B^C^  similar. 

Proof.         Take  AE^  A'B',  and  AII=  A'C. 

Draw  EH. 
Then  from  the  given  proportion, 

AB^AC_  i 

ae'  ah 

.-.  EHis  II  to  BC,  §  312 

{if  a  line  divide  two  sides  of  a  A  proportionally,  it  is  II  to  the  third  side). 
Hence  in  the  A  ABC &nd  AEH  '  "^ 

Z  ABO=  Z.  AEH,  §  106 

and  ZACB  =  ZAHE, 

{being  ext.-int.  AofW  lines). 

.-.  A  ABO  siud  AEH  are  similar,  §  322 

{two  A  are  similar  if  two  A  of  one  are  equal  respectively  to  two  A  of  the 

other). 

.'.  AB:  AE  ^  BO:  EH; 

that  is,  AB  :  A'B'  =  BO:  EH. 


SIMILAR    TRIANGLES. 


147 


But  by  hypothesis, 

The  last  two  proportions  have  the  first  three  terms  equal 
each  to  each  ;  therefore  the  fourth  terms  are  equal ;  that  is, 

ub:=  b'C. 

Hence  in  the  A  ^^.fi'and  A'B'C, 

EE:=B'C\  AE=A'B\  and^^=^'C'. 

.-.  A  AEB:=  a  A'B'C\  §  160 

{having  three  sides  of  the  one  equal  respectively  to  three  sides  of  the  other). 

But  A  AEH  is  similar  to  A  ABC. 

.-.  A  A'B^C  is  similar  to  A  ABC.  ^  e.  d. 

325.  Scholium.  The  primary  idea  of  similarity  is  likeness 
u/form ;  and  the  two  conditions  necessary  to  similarity  are  : 

I.  For  every  angle  in  one  of  the  figures  there  must  be  an 
equal  angle  in  the  other,  and 

II.  The  homologous  sides  must  be  in  proportion. 

In  the  case  of  triangles,  either  condition  involves  the  other, 
but  in  the  case  of  other  polygons,  it  does  not  follow  that  it  one 
condition  exist  the  other  does  also. 


Q' 


Thus  in  the  quadrilaterals  Q  and  Q\  the  homologous  sides 
are  proportional,  but  the  homologous  angles  are  not  equal. 

In  the  quadrilaterals  R  and  B!  the  homologous  angles  are 
equal,  but  the  sides  are  not  proportional. 


148  PLANE    GEOMETRY. BOOK    III. 


Pkoposition  VII.     Theorem. 

326.  If  two  triangles  have  an  angle  of  the  one  equal 
to  an  angle  of  the  other,  and  the  including  sides  pro- 
portional, they  are  similar. 


In  the  triangles  ABC  and  A'B'C\  let  ZA  =  ZA',  and 

AB  ^  AC 
A'B'      A'C' 

To  prove  A  ABC  and  A'B'C  similar. 

Proof.    Apply  the  A  A'B'O'  to  the  A  ABC,  so  that  Z  .4' 
diall  coincide  with  Z  A. 

Then  the  A  A'B'C  will  take  the  position  of  A  A  EH. 


AB  _  AC 
A'B'     A'C 


^T  Jil^  Ji.U  XT 


That  is,  ^=  A£- 

—  AE     AH 

Therefore  the  line  UH  divides  the  sides  AB  and  ^C  pro- 
portionally ; 

.-.  ^^is  II  to  BC  §  312 

{if  a  line  divide  two  sides  of  a  A  proportionally,  it  is  II  to  the  third  side). 

Hence  the  A  ABC  and  AEH  are  mutually  equiangular 
and  similar. 

.-.  A  A'B'C  is  similar  to  A  A  BC. 

Q.  E.  D. 


SIMILAR   TRIANGLES. 


149 


Proposition  VIII.     Theorem. 

327.  If  two  triangles  have  their  sides  respectively 
parallel,  or  respectively  perpendicul/ir,  they  are  siin- 
ilar, 

A 


In  the  triangles  A'B'C  and  ABC  let  A'B',  A'C,  B'C  be 
respectively  parallel,  or  respectively  perpendicular, 
to  AD,  AC,  BC. 

To  prove         A  A'B'C  and  ABC  similar. 

Proof.  The  corresponding  A  are  either  equal  or  supplements 
..f  each  other,  §§112,113 

(if  two  A  have  their  sides  II,  or  JL,  they  are  equal  or  supplementary). 
Hence  we  may  make  three  suppositions : 

1st.  A  +  A'  =  2vtA,   B  +  B'  =  2rtA,    C+Cf==2rtA. 
2d.  A  =  A',  B-}-B'  =  2rt.A,    C+C"=2rt.Zs. 

3d.  A=^A\  B  =  B\      .'.  C=  C\  §  140 

Since  the  sum  of  the  A  of  the  two  A  cannot  exceed  four 
right  angles,  the  third  supposition  only  is  admissible.        §  138 

/.  the  two  A  ^5(7  and  A'B'C*  are  similar.       §  321 

{tiuo  mutually  emdangular  ^  are  similar). 

a  E.  o 


150 


PLANE    GEOMETRY. 


BOOK    III. 


Proposition  IX.     Theorem. 

328.  The  homologous  altitudes  of  two  similar  tri- 
angles  have  the  saine  ratio  as  any  two  homologous 
sides. 


A  o 


uV    0 


In  the  two  similar  triangles  ABC  and  A'B'C,  let  the 
altitudes  be  CO  and  CO'. 


To  prove 


CO 


AC       AB 


CO'     A'C     A'£' 

Proof.    In  the  rt.  A  COA  and  CO' A', 

A  A^Z.  A\  %  319 

{being  homologous  A  of  the  similar  A  ABC  and  A'B'C). 

.-.A  COA  and  CO' A'  are  similar,  s  323 

{two  rt.  A  having  an  acute  Z  of  the  one  equal  to  an  acute  Z  of  the  other 
are  similar), 

*  *  ca    A'c' 

In  the  similar  A  ^5(7  and  A^B'C\ 

AC  _  AB 
A'C'    A'B' 


§319 


Therefore, 


CO  _  AC  ^  AB 
CO'      A'C      A'B' 


Q.  E.  D. 


SIMILAR   TRIANGLES.  151 


Proposition  X.     Theorem. 


329.  Straight  lines  d^awn  through  flie  sajne  point 
intercept  proportional  segments  upon  two  parallels. 


Let  the  two  parallels  AE  and  A'E'  cut  the  straight 
lines  OA,  OB,  OC,  OD,  and  OE. 

AB       BC       CD       DE 


Topi 


A'B'     BC     CD'     D^I? 


Ax.  1 


Proof.  Since  A'E'  is  II  to  AE,  the  pairs  of  A  OAB  and 
OA'B\  OBC  and  OB^C,  etc.,  are  mutually  equiangular  and 
similar, 

.    AB       OB  ^.    BC       OB  goio 

{homologous  sides  of  similar  A  are  proportional). 

•  •  A'B'     B'C' 

In  a  similar  way  it  may  be  shown  that 

BC  _  CD        .  CD  _  DE 
B'C      CD'  ^^    CD'     D'E'' 

Q.  E.  D 

Remark,    A  condensed  form  of  writing  the  above  is 

A^=.(OB^\^    BC      f  00\  _   CD    ^  fO^\-  DE 
A'Bf      \OB'J      B'C     [OCJ      CI/     \0b')     D'E'^ 

where  a  parenthesis  about  a  ratio  signifies  that  this  ratio  is  used  to 
prove  the  equality  of  the  ratios  immediately  preceding  and  following  it- 


152 


PLANE   CtEOMETRY 


BOOK  in. 


Proposition  XI.     Theorem. 

330.  Conversely  :  //  three  '  or  more  non-parallel 
straight  lines  infercej)t  proportional  segments  upon 
two  parallels,  they  pass  through  a  coTninon  point. 

0 


BL 


J) 


IF 


Let  AB,  CD,  EF,  cut  the  parallels  AE  and  BF  so  that 
AC  :  BD=GE  '.  DF. 

To  prove  that  AB,  CD,  EF  prolonged  meet  in  a  point. 

Proof.    Prolong  AB  and  CD  until  they  meet  in  0. 

Join  OE. 

If  we  designate  by  F  the  point  where  OE  cuts  BF,  we 
shall  have  by  §  329, 

AC:BD=CE'.DF\ 
But  by  hypothesis 

AC:BD=CE:DF. 

These  proportions  have  the  first  three  terms  equal,  each  to 
each ;  therefore  the  fourth  terms  are  equal ;  that  is, 

DF^DF, 

.'.  F'  coincides  with  F. 

.'.  ^i*^  prolonged  passes  through  O. 

.'.  AB,  CD,  and  ^i^ prolonged  meet  in  the  point  0. 

"-^  Q.  E.  Q 


SIMILAR   POLYGONS. 


153 


Similar  Polygons. 
Proposition  XII.     Theorem. 

331.   If  two  polygons  are  composed  of  the  sniy\j>,  nurn' 
ber  of  triangles,  similar  each  to  eaxih,  arbd  similarly 
placed,  the  polygons  are  similar. 
E 


B  C  B' 

In  the  two  polygons  ABODE  and  A'^'CU'E',  let  the 
triangles  AEB,  BEC,  CED  be  similar  respectively  to 
the  triangles  A'E'B',  B'E'C,  C'E'D'. 

To  prove       ABODE  similar  to  A'B'C'D'E. 

Proof.  Z.A--=Z.A\  §319 

{being  homologotis  zi  of  similar  A). 

/.ABE-=/.A'B'E'  §319 

Z  EBC  =  Z  EB'C 


Also, 
and 

By  adding, 


Z  ABC  =  Z  A'B'C. 
In  like  manner  we  may  prove  Z  BCD  —  Z  B'C'D',  etc. 
Hence  the  two  polygons  are  mutually  equiangular. 
Now 

AE  __  AB  _f  EB\_  BC^f  EC\       CD 
B'C     \ 


A'E' 


ED 


E'D' 


A'B'     \E'B'J     B'C     \E'C'J     CD' 

{the  homologous  sides  of  similar  A  are  proportional). 

Hence  the  homologous  sides  of  the  polygons  are  proportional. 

Therefore  the  polygons  are  similar,  §.319 

(having  their  homologous  A  equal,  and  their  homologous  sides  proportional). 


164 


PLANE   GEOMETRY.  —  BOOK    III. 


Proposition  XIII.     Theorem. 

332.  If  two  -polygons  are  siinilar,  they  are  composed 
of  the  same  rvwinher  of  triangles,  similar  each  to  each, 
and,  similarly  plax^ed. 


B  C  B'  C 

Let  the  polygons  ABODE  and  A'B'C'D'E'  be  similar. 

From  two  homologous  vertices,  as  U  and  U',  draw  diagonals 
jEB,  EC,  and  E'B\  E^CK 

To  prove  A  EAB,  EBC,  ECD 

similar  respectively  to  A  E'A'B',  E'B'C,  E'C'I)'. 
Proof.    In  the  A  EAB  and  E'A'B',. 
ZA  =  ZA', 
{being  homologous  A  of  similar  polygons) ; 

AE  __  AB 
A'E' 


and 


§319 
§  319 


A'B' 

{being  homologous  sides  of  similar  polygons). 
.'.  A  EAB  and  E'A'B'  are  similar,  §  326 

(having  an  Z  of  the  one  equal  to  an  Z  of  the  other,  and  the  including  sides 
proportional). 

Also,  ZABC=ZA'B'C',  (1) 

{being  homologous  A  of  similar  polygons). 
And  ZABE--=ZA'B'E',  (2) 

{being  homologous  A  of  similar  A), 

Subtract  (2)  from  (1), 

ZEBC=Z  E'B'C.  Ax.  3 


Now 


And 


SIMILAR    POLYGONS. 

EB  _  AB 
E'B'     A'B'^ 

(being  homologous  sides  of  similar  ^). 

BO  _  AB 
B'C     A'B'' 

{being  homologous  sides  of  similar  polygons). 

.    EB       BO 


155 


Ax.  1 
§326 


'  •  E'B'     B'O* 

.-.  A  EBO ?ind  E'B'C*  are  similar, 

{having  an  /.  oj  the  one  equal  to  an  Z  of  the  other,  and  the  including  sides 

proportional). 

In  like  manner  we  may  prove  A  EOD  and  E^O^D^  similar. 

a  E.  D. 

Proposition  XIV.     Theorem. 

333.  The  perimeters  of  two  similar  polygons  have 
the  same  ratio  as  any  two  homologous  sides. 
E 


B  C  B'  C 

Let  the  two  similar  polygons  be  ABODE  and  A'B'C  D'E', 
and  let  P  and  F*  represent  their  perimeters. 

To  prove  F:F'  =  AlB:A'B'. 

AB  :  A'B'  =  BO:  B'O'  -  OB  :  O'D',  etc.,        §  319 
{the  homologous  sides  of  similar  polygons  are  proportional). 

.-.  AB  +  BO,  etc. :  A'B'  +  B'O',  etc.  =  AB :  A'B',  §  303 

{in  a  series  of  equal  ratios  the  sum  of  the  antecedents  is  to  the  sum  of  the 
consequents  as  any  antecedent  is  to  its  consequent). 

That  is,  F:P  =  AB:A'Br  Q.E,a 


156 


PLANE    GEOMETRY.  —  BOOK    III. 


Numerical  Properties  of  Figures. 
Proposition  XV.     Theorem. 

334.  If  in  a  right  triangle  a  perpendicular  is  drawn 
from  the  vertex  of  the  right  angle  to  the  hypotenuse  : 

I.  The  perpendicular  is  a  mean  proportional  be- 
tween the  segments  of  the  hypotenuse. 

II.  Each  leg  of  the  right  triangle  is  a  mean  pro- 
portional between  the  hypotenuse  and  Us  adjacent 
segment. 


In  the  right  triangle  ABC,  let  BF  be  drawn  from  the 
vertex  of  the  right  angle  B,  perpendicular  to  AC. 
I.    To  prove  AF:  BF=  BF:  FC. 

Proof.    In  the  rt.  A  ^^i^and  BAC 

the  acute  Z.  A  is  common. 
Hence  the  A  are  similar.  §  323 

in  the  rt.  A  ^CT^and  BCA 

the  acute  A  C\s  common. 
Hence  the  A  are  similar.  §  323 

Now  as  the  rt.  A  ^-Si^and  CBFsltb  both  similar  to  ABC, 
they  are  similar  to  each  other. 

In  the  similar  A  ^^i^and  CBF, 

AF,  the  shortest  side  of  the  one, 
BF,  the  shortest  side  of  the  other, 
BF,  the  medium  side  of  the  one, 
FC,  the  medium  side  of  the  other.  • 


II. 

and 


To  prove 


ACiAB 
AC:  BO 


AB:AF, 
BC  :  FC. 


NUMERICAL    PROPERTIES   OF    FIGURES.  157 

In  the  similar  A  ABC  and  ABF, 

AC,  the  longest  side  of  the  one, 
AB,  the  longest  side  of  the  other, 

AB,  the  shortest  side  of  the  one, 
AF,  the  shortest  side  of  the  other. 

Also  in  the  similar  A  ^J5Cand  FBC, 

AC,  the  longest  side  of  the  one, 

:  BC^  the  longest  side  of  the  other, 
: :  BC,  the  medium  side  of  the  one, 
:  FC,  the  medium  side  of  the  other.  q.  e.  o. 

335.  Cor.  1.  The  squares  of  the  two  legs  of  a  rif/ht  triangle 
are  proportional  to  the  adjacent  segments  of  the  hypotenuse. 

The  proportions  in  II.  give,  by  §  295, 

AB'^ACxAF,  and  BC'  =  ACx  CF. 
By  dividing  one  by  the  other,  we  have 
A^_ACxAF_AF 
-BC^     ACxCF      CF' 

336.  CoR.  2.  The  squares  of  the  hypotenuse  and  either  leg 
are  propo7'iional  to  the  hypotenuse  and  adjacent  segment. 

^  AC"  _ACxAC  _AC 

°''  AB"     ACxAF     AF 

337.  Cor.  3.  An  angle  inscribed  in  a  semicircle  is  a  right 
angle  (§  264).  Therefore, 

I.  The  perpendicular  from  aiiy  point  in 
the  circumference  to  the  diameter  of  a  circle 
is  a  mean  proportional  between  the  segments 
of  the  diameter.  "^      ^^ 

II.  The  chord  drawn  from  the  point  to  eithe)'  extremity  of  the 
diamete)'  is  a  mean  proportional  between  the  diameter  and  the 
adjacent  segment. 

Remark.  The  pairs  of  correspoftding  sides  in  similar  triangles  may  be 
called  longest,  shortest,  medium,  to  enable  the  beginner  to  see  quickly 
these  pairs ;  but  he  must  not  forget  that  two  sides  are  homologous,  not 
because  they  appear  to  be  the  longest  or  the  shortest  sides,  but  because 
they  lie  opposite  corresponding  equal  angles. 


158 


PLANE    GEOMETRY, 


BOOK    III. 


4 


Proposition  XVI.     Theorem. 


338.  The  sum  of  the  squares  of  the  two  legs  of  a  right 
triangle  is  equal  to  the  square  of  the  hypotenuse. 


Let  ABC  be  a  right  triangle  with  its  right  angle  at  B. 

To  prove  AB'  +  BC'  =  AC\ 

Proof.  Draw  BF 1.  to  AC. 

Then  AF  =  AC x  AF  §  334 

and  BC"  =  ACx  CF 

By  adding,  AB"  +  BC"  ^-  AC(AF-j-  CF)  =  AC\        q.  e.  d. 

339.  Cor.    The  square  of  either  leg  of  a  right  triangle  is  equal 
to  the  difference  of  the  squares  of  the  hypotenuse  and  the  other  leg. 

340.  Scholium.  The  ratio  of  the  diagonal  of  a 
square  to  the  side  is  the  incommensurable  num- 
ber V2.  For  if  ^  C  is  the  diagonal  of  the  square 
ABCD,  then 

AB'  +  BC\  or  IC"  =  2  AB\ 


AC 


Divide  by  AB  ,  we  have 


AC' 


'Z,  or 


AC 


V2. 


AB'  ^B 

Since  the  square  root  of  2  is  incommensurable,  the  diagonal 
and  side  of  a  square  are  two  incommensurable  lines. 

341,    The  projection  of  a  line  CD  upon  a  straight  line  ^^  is 
that  part  of  the  line  AB  comprised  q 

between  the  perpendiculars  CP  and 
DR  let  fall  from  the  extremities  of 
CD.  Thus,  PR  is  the  projection  of 
CD  upon  AB.  ''    F  R 


A- 


'B 


NUMERICAL    PROPERTIES    OF    FIGURES.  159 


Proposition  XVII.     Theorem. 

342.  In  any  triangle,  the  square  of  the  side  opposite 
an  acute  angle  is  equal  to  the  sum  of  the  squares  of 
the  other  two  sides  diminished  hy  twice  the  product 
of  one  of  those  sides  and  the  prqjection  of  tJie  other 
upon  that  side. 


Let  C  be  an  acute  angle  of  the  triangle  ABC,  and 
DC  the  projection  of  AC  upon  BC. 

To  prove      AB'  -  BC''-{-  AC"-2BCx  Da 
Proof.    If  D  fall  upon  the  base  (Fig.  1), 

I)B  =  BC-I)C; 
If  B  fall  upon  the  base  produced  (Fig.  2), 

nB=^BC-Ba 

In  either  case, 

BB'  ^  BC'-{-  BC'  -2B0x  BC. 

Add  AB^  to  both  sides  of  this  equality,  and  we  have 

AB"  +  BB". ^  BC'+ 10 ■\-  BC'-  2BCx  BC} 

But  ^^c{-BB'  =  AB'^,^  §338 

and  A^+^'^AC', 

(the  Sinn  of  the  squares  ofihe  two  legs  of  a  rt.  A  is  equal  to  the  square 
of  the  hypotenuse). 

Put  AB^  and  AC'  for  their  equals  in  the  above  equality, 
AB'  -  BC''  +  AC"-  2BCx  BC. 


160 


PLANE    GEOMETRY, 


BOOK    III. 


Proposition  XVIII.     Theorem. 

343.  In  any  obtuse  triangle,  the  square  of  the  side 
opposite  the  obtuse  angle  is  equal  to  the  sum  of  the 
squares  of  the  other  two  sides  increased  by  twice  the 
product  of  one  of  those  sides  and  the  projection  of 
the  other  upon  that  side. 


Let  C  be  the  obtuse  angle  of  the  triangle  ABC,  and 
CD  be  the  projection  of  AG  upon  BC  produced. 

To  prove      AB' =  BC'' -\- AO^ -^2BCx  DC. 

Proof.  DB=^BC-^  DC. 

Squaring,    DB\^  BC''  +  DO"  -\-2BCx  DC      . 

Add  AD  to  both  sides,  and  we  have 

Aff  +  DB'  =  BC'-\-AD'+DC"-^2BC  x  DC 

But  AD'  +  DF^AB",  §338 

and  AD'+D0'  =  AG\ 

(the  sum  of.  the  squares  of  the  two  legs  of  a  rt.  A  is  equal  to  the  square 
of  the  hypotenuse).: 

Put  AB  and  AC   for  their  equals  in  the  above  equality, 

AB'=:^BC'  +  AG'  +  2BCxDC 

a  E.  o. 

Note.    The  last  three  theorems  enable  us  to  compute  the  lengths  of 
the  altitudes  if  the  lengths  of  the  three  sides  of  a  triangle  are  known. 


NUMERICAL    PROPERTIES   OF    FIGURES.  161 


Proposition  XIX.     Theorem. 

344.  I.  The  sum  of  the  squares  of  two  sides  of  a  tri- 
angle is  equal  to  twice  the  square  of  half  the  third 
side  increased  by  twice  the  square  of  the  median  upon 
that  side. 

II.  The  difference  of  the  squares  of  two  sides  of  a 
triangle  is  equal  to  twice  the  -product  of  the  third 
side  hy  the  projection  of  the  median  upon  that  side. 

A 


In  the  triangle  ABC  let  A }f  be  the  median,  and  MD 
the  projection  of  AM  upon  the  side.  BC.  Also  let  AB 
be  greater  than  AC. 

Toprove       I.  AF  +  AC^  =  2BM'  -{-  2AM^. 
II.  AB"  ~  AC"  =  2£Cx  MD. 

Proof.  Since  AB>AC,  tlie  Z.  AMB  will  be  obtuse,  and 
the  Z  AMC  will  be  acute.  .  §  153 

Then        AB"  =  Blf\  +  AM'  +  2 BM X  MB,  §  343 

{in  any  obtuse  A  the  square  of  the  side  opposite  the  obtuse  Z  is  equal  to  the 
sum  of  the  squares  of  the  other  two  sides  increased  by  twice  the  product 
of  one  of  those  sides  and  the  projection  of  the  other  on  that  side) ; 

and  AC"  =  MC'  +  -Sf '  -  2 MCx  MD,  §  342 

{in  any  A  the  square  of  the  side  opposite  an  acute  Z  is  equal  to  the  sum  of 
the  squares  of  the  other  two  sides  diminished  by  twice  the  product  of  one 
of  those  sides  and  the  projection  of  the  other  upon  that  side). 

Add  these  two  equalities,  and  observe  that  BM=  3fC. 
Then  AB"  +  AO"  =  2BM\-h2AM\ 

Subtract  the  second  equality  from  the  first. 
Then  AB"  -  AC' =  2BCx  MD.  q.e.d. 

Note.  This  theorem  enables  us  to  compute  the  lengths  of  the  mediam< 
if  the  lengths  of  the  three  sides  of  the  triangle  are  known. 


162  PLANE  GEOMETRY.  —  BOOK   III. 


Proposition  XX.     Theorem. 

345.  If  any  chord  is  drawn  through  a  fixed  point 
within  a  circle,  the  product  of  its  segments  is  con^ 
stant  in  whatever  direction  the  chord  is  drawn. 


iLet  any  two  chords  AB  and  CD  intersect  at  0. 

To  prove  OA  X  OB  =  OD  X  00. 

Proof.  Draw  ^C  and  ^i>. 

In  the  ^  AOO  and  BOD, 

AQ^AB,  §263 

(eacA  heing  measured  by  ^  arc  AD). 

ZA  =  ZIl,  §263 

(each  being  measured  by  ^  arc  BO). 

.'.  the  A  are  similar,  §  322 

(two  A  are  similar  when  two  A  of  the  one  are  equal  to  two  A  of  the  other) 

Whence       OA,  the  longest  side  of  the  one, 

:  OB,  the  longest  side  of  the  other, 

:  .  00,   the  shortest- side  of  the  one, 

:  OB,  the  shortest  side  of  the  other. 

.-.  OAxOB^OBx  00.  §  295 

aE.  D 
346.    Scholium.     This  proportion  may  be  written 

OA^qO        OA^J_, 

OB      OB'  ^^  OB      OB' 
\  00 

that  is,  the  ratio  of  two  corresponding  segments  is  equal  to 
the  reciprocal  of  the  ratio  of  the  other  two  corresponding 
segments.  In  this  cast  the  segments  are  said  to  be  reciprocally 
proportional. 


NUMERICAL    PROPERTIES    OF    FIGURES. 


163 


Proposition  XXI.     Theorem. 

347.  If  from  a  fijced  point  without  a  circle  a  secant 

is  drawn,  ths  product  of  the  secant  and  its  external 

segment  is  constant  in  whatever  direction  the  secant 

is  drawn, 

O 


Let  OA  and  OB  be  two  secants  drawn  from  point  0. 

To  prove  OAxOO=OBx  01). 

Proof.  Draw  J5C  smd  AD. 

In  the  A  OAD  and  OBC 

Z  0  is  common, 

ZA  =  Z.B,  §263 

{each  being  measured  by  J  arc  CD). 

.'.  the  two  A  are  similar,  §  322 

{two  i^  are  similar  when  two  A  of  the  one  are  equal  to  two  A  of  the  other). 

Whence         OA,  the  longest  side  of  the  one, 
OB,  the  longest  side  of  the  other, 
OD,  the  shortest  side  of  the  one, 
OC,  the  shortest  side  of  the  other. 

.-.  OA  X  00=  OB  X  OD.  §  295 

Q.  E.  D. 

Remahk.  The  above  proportion  continues  true  if  the  secant  OB  turns 
about  0  until  B  and  D  approach  each  other  indefinitely.  Therefore,  by 
the  theory  of  limits,  it  is  true  when  B  and  D  coincide  at  H.  Whence, 
OA  X  0C=^  0H\ 

This  truth  is  demonstrated  directly  in  the  next  theorem. 


164  PLANE    aEOMETRY.  —  BOOK    III. 


Proposition  XXII.     Theorem. 

348.  If  from  a  point  without  a  circle  a  secant  and 
a  tangent  are  drawn,  the  tangent  is  a  meaTi  propor- 
tional between  the  whole  secant  and  the  external 
segment. 


Let  OB  be  a  tangent  and  OC  a  secant  drawn  from 
the  point  0  to  the  circle  MBC. 

To  prove  00 :  OB  =  OB:  OM. 

Proof.  Draw  ^ Jf  and  ^a 

In  the  A  Oj^Jf  and  OBC 

^  0  is  common. 

Z  OBM  is  measured  by  ^  arc  3fB,  §  269 

(being  an  Z  formed  hy  a  tangent  and  a  chord). 

Z  C  is  measured  by  ^  arc  BM,  §  263 

(being  an  inscribed  £). 

,\/.OBM=Z.C. 

.-.  A  0^(7 and  OBM ^xq  similar,  §  322 

(having  two  A  of  the  one  equal  to  two  A  of  the  other). 

Whence         00,    the  longest  side  of  the  one, 
OB,  the  longest  side  of  the  other, 
OB,  the  shortest  side  of  the  one, 
OM,  the  shortest  side  of  the  other. 

Q.  E.  D. 


V 


NUMERICAL    PROI-EJEITIES    OF    FIGURES.  165 

Proposition  XXIII.     Theorem. 

349.  The  square  of  the  bisector  of  an  angle  of  a 
triangle  is  equal  to  the  product  of  the  sides  of  this 
angle  diminished  by  the  product  of  the  segments 
determined  by  the  bisector  upon  the  third  side  of  the 
triangle. 


'K- 


m 


E 
Let  AD  bisect  the  angle  BAC  of  the  triangle  ABC. 

To  prove        Iff  =  ABx  AC-  DB  X  DC. 

Proof.      Circumscribe  the^O  ^jBC  about  the  A  ABC. 

Produce  AD  \o  meet  the  circumference  in  Ey  and  draw  EG. 

Then  in  the  A  ABD  and  AEC, 

ZBAD  =  ZCAE,  Hyp. 

ZB  =  Z  E,  §  263 

{each  being  measured  Sy  \  the  arc  AC). 

.-.  A  ABD  and  AEC  sltb  similar,  §  322 

[tivo  ^  are  similar  if  two  A  of  the  one  are  equal  respectively  to  two  A  oj 

the  other). 

Whence        AB,  the  longest  side  of  the  one, 

AE,  the  longest  side  of  the  other, 

AD,  the  shortest  side  of  the  one, 

AC,  the  shortest  side  of  the  other. 

.\  AB  X  AC=  AD  X  AE^  §  295 

=  AD(AD+DE) 

=  AD"  +  AD  X  DE. 

But  AD  X  DE=  DB  x  DC,  §  345 

(the  prodjict  of  the  segments  of  a  chord  drawn  through  a  fixed  point  in 
a  Q  is  constant). 

.'.  ABx  AC=  AD'  +  DBxDC. 

Whence        AD"  =  ABxAC—  DB  x  DC.  q.  e.  d. 

Note.   This  theorem  enables  us  to  compute  the  lengths  of  the  bisectors 
of  the  angles  of  a  triangle  if  the  lengths  of  the  sides  are  known. 


166 


PLANE    GEOMETRY.  —  BOOK    III. 


Proposition  XXIV.     Theorem. 

350.  In  any  triangle  the  product  of  two  sides  is 
equal  to  the  product  of  the  diameter  of  the  circurn- 
scrihed  circle  by  the  altitude  upon  the  third  side. 


Let  ABC  he  a   triangle,  AD  the   altitude,  and  ABC 
the  circle  circumscribed  about  the  triangle  ABC. 

Draw  the  diameter  AE,  and  draw  EC. 


To  prove 

ABxAC=AExAD. 

,,' 

Proof.    In 

the  A  ABD  and  AEC, 

/.  BDA  is  a  rt.  Z, 

Cons. 

Z.  EC  A  is  a  rt.  Z, 

§264 

(being  inscribed  in  a  semicircle), 

2.udZB=^ZE. 

§263 

.-.A  ABB  and  ^^Care  similar, 

§323 

wo  rt.  A  hav 

inq  an  acute  Z  of  the  one  equal  to  an  acute  Z  of 

the  other 

w 


are  similar). 
AB,  the  longest  side  of  the  one, 
AE,  the- longest  side  of  the  other, 
AB,  the  shortest  side  of  the  one, 
A  Cf,  the  shortest  side  of  the  other. 
".-.  ABxAC=AExAB. 


§295 

Q.  E.  D. 


Note.  This  theorem  enables  us  to  compute  the  length  of  the  radius  of 
a  circle  circumscribed  about  a  triangle,  if  the  lengths  of  the  three  sides 
of  the  triangle  are  known. 


i'KuBLEMS    OF    CONSTKUCTION.  167 

Problems  of  Construction. 

Proposition  XXV.     Problem. 

351.  To  divide  a  given  straight  line  into  parts  pro- 
portional to  any  number  of  given  lines. 

A, IL A'     B 


n  ■■■•..  \ 


m- 


p. .  ^ 

Let  AB,  m,  n,  and  p,  be  given  straight  lines. 
To  divide  AB  into  parts  proportixynal  to  m,  w,  and  p. 

Construction.     Draw  AX,  making  an  acute  Z  with  AB. 

On  ^Xtake  AC=m,  CE=n,  EX=p. 

Draw  BX. 

From  ^and  Cdraw  ^^and  CH  II  to  BX. 

uff'and  JTare  the  division  points  required. 

Proof,  (AE\^AB^IJK_^KB_^  ^  .^^ 

\AE)      AC      CE      EX  ^ 

(a  line  drawn  through  two  sides  of  a  A  II  to  the  third  side  divides  those 
sides  proportionally). 

.-.  AJI  :  UK  :  KB  =  AC  :  CU  :  EX. 

Substitute  m,  n,  and  p  for  theii-  equals  AC,  CE^  and  EX. 

Then  AH  :  UK  :  KB  =  m  :  n  :p. 

Q.E.F 


168  PLANE   GEOMETRY.  —  BOOK    III. 


5 

Proposition  XXVI.     Problem.  < 

352.  To  find  a  fourth  -proportional  to  three  given 
straight  lines,  t-^^ 


m B      n       C  « 


f: 


Let  the  three  given  lines  be  m,  n,  and  p. 

To  find  a  fourth  'proportional  to  m,  n,  and  p. 

Draw  Ax  and  Ay  containing  any  acute  angle. 

Oonstruction.  On  Ax  take  AB  equal  to  m,  BG=  n. 

On  Ay  take  AD  =  p. 

Draw  BD. 

From  Cdraw  OF  \\  to  BD,  to  meet  Ay  at  F. 

DF  is  the  fourth  proportional  required. 

Proof.  AB  :  BC=  AD:  DF,  §  309 

(a  line  drawn  through  two  sides  of  a  A  W  to  the  third  side  divides  those 
sides  proportionally). 

Substitute  m,  ?i,  snidp  for  their  equals  AB,  BO,  and  AD. 

Then  m  '.  n=^  p  :  DF. 


PLOBLEMS   OF    CONSTRUCTION.  169 


Proposition  XXVII.     Problem. 

353.  To   -find    a    third    proportional   to    two   given 
straight  lines. 

A 


T>^ - - -----E 

Let  m  and  n  be  the  two  given  straight  lines. 
To  find  a  third  proportional  to  m  and  n. 
Oonstmction.     Construct  any  acute  angle  A, 
and  take  AB=^m.,  AC=n. 
;^oduce  AB  to  D,  making  BB  =■-  AC. 
Join  BC. 
Through  D  draw  DE  II  to  BO  to  meet  reproduced  at  E. 
CE  is  the  third  proportional  to  AB  and  AC. 

Proof.         *  AB:  BB  =  AC:  CE.  §  309 

(a  line  drawn  through  two  sides  of  a  A  W  to  the  third  side  divides  those 
sides  proportionally). 

Substitute,  in  the  above  proportion,  ^Cfor  its  equal  BI). 

Then   AB  :  AC=AC:  CE. 

That  is,  m:n  —  n  :  CE. 

aE.F. 


Ex.  217.   Construct  x,  if  (1)  x  =  — ,  (2)  or  =  -• 

c  c 

Special  Cases  :  (1)  a  =  2,  6  =  3,   c  =  4  ;    (2)  a  =  3,   &  =  7,    c  =  11 ;  (3) 

a  =  2,  c  -=  3  ;   (4)  a  =  3,  c  =  5 :   (5)  a  -  2c. 


170  PLANE    GEOMETRY.  —  BOOK    III. 


Proposition  XXVIII.     Problem. 

354.  To  -find,  a  -mean  proportioiial  between  two  given 
straight  lines. 

II 


m 


^  m         c  ^  Ji 


-E 


Let  the  two  given  lines  be  m  and  n. 

To  find  a  mean  proportional  between  m  and  n. 
Construction.        On  the  straight  line  AE 

take  AC^^  m,  and  CB  =  n. 
On  AB  as  a  diameter  describe  a  semi-circumference. 
At  C  erect  the  _L  CH  to  meet  the  circumference  at  H. 
CITis  a  mean  proportional  between  m  and  71. 

Proof.  .-.  AC  -CH  =CH  :  CB,  §  337 

[the,  ±  let  fall  from  a  point  in  a  circumference  to  the  diaineter  of  a  circle 
is  a  mean  proportional  between  the  segments  of  the  diameter). 

Substitute  for  ^(7  and  CB  their  equals  m  and  71. 

Then  m  :  CII  ^  CB  \  n. 

Q.  E.  F. 

355.  A  straight  lin«  is  divided  in  extre7ne  and  7nean  ratio, 
when  one  of  the  segments  is  a  mean  proportional  between  the 
whole  line  and  the  other  segment. 


Ex.218.    Construct  a?  if  a;  ="N/a6. 

Special  Cases  :  (1)  a  =  2,  6  -  3  ;  (2)  a  =  1,  6  =  5  ;  (3)  a  =  3,  6  =  7. 


PROBLEMS   OF   CONSTRUCTION. 


171 


Proposition  XXIX.     Problem. 

356.  To  divide  a  given  line  in  extreme  and  mean 
ratio. 


\Q 


E.-" 


O- 


A  C  B 

Let  AB  be  the  given  line. 
To  divide  AB  in  extreme  and  mean  ratio. 
Oonstniction.     At  B  erect  a  ±  BE  equal  to  one-half  of  AB. 
From  _E'as  a  centre,  with  a  radius  equal  to  EB,  describe  a  O. 
Draw  AE,  meeting  the  circumference  in  i^and  G. 
On  ^^take^C=^i^. 
On  BA  produced  take  AC  ^  AG. 
Then  AB  is  divided  internally  at  C  and  externally  at  C" 
in  extreme  and  mean  ratio. 

Proof.  AG  :  AB  =  AB '.  AF,  §  348 

{if  froin  a  point  without  a  O  a  secant  and  a  tangent  are  drawn,  the  tan- 
gent is  a  mean  proportional  between  the  whole  secant  and  the  external 
segment). 

Then  by  §  301  and  §  300, 

AG^  AB :  AB  =  AB  -  AF  :  AF,  (J) 

AG  +  AB:AG  =  AB  +  AF  :  AB.  (2J 

By  construction    FG  =  2EB=^  AB. 

.-.  AG~AB  =  AG-FG  =  AF=Aa 

Hence  (1)  becomes 

AC  :AB  =  BC  :AC^. 

or,  by  inversion,      AB  :   AC=  AC  :  BCJ ^ 

Again,  since       C'A  =  AG  =  AB  +  Al\  ^ 

(2)  becomes  C'B  :  C'A  =  C'A  :  AB.  I  ^ 

Q.  E.  F. 


172 


PLANE    GEOMETKY.  —  BOOK    III. 


Propositioit  XXX.     Problem. 

357.    Upon  a  given  line  Jiomologous  to  a  given  side 
of  a  given  polygon,  to  construct  a  polygon  similar  to 
the  given  polygon. 
E 


Let  A'E'  be  the  given  line  homologous  to  AE  of  the 
given  polygon  ABODE. 

To  construct  on  A'U'  a  jpolygon  similar  to  the  given  polygon. 

Oonstmction.    From  E  draw  the  diagonals  EB  and  EC. 

From  E'  draw  E'B\  E'C\  and  E' D\ 

m^VmgAA'E'B\  B'E'C\  and  C'E'D'  equal  respectively  to 

A  AEB,  BEC,  and  CED. 

From  A'  draw  A'B\  making  Z.  E'A'B'=  Z  EAB, 

and  meeting  E^B^  at  B\ 
From  B'  draw  B'C\  making  Z  E^BW'  =  Z  EBC, 

and  meeting  E'C  at  C. 

From  C  draw  O^B',  making  Z  E'C'B'  =  Z  ECB, 

and  meeting  E'B'  at  I)'. 

Then  A'B'C'B'E'  is  the  required  polygon. 

Proof.    The  corresponding  A  ABE  and  A'B'E\  EBC  and 
E'B'C,  ECD  and  E'C'H  are  similar,  §  322 

{two  A  ar&  similar  if  they  have  two  A  of  the  one  equal  respectively  to  two 
A  of  the  other). 

Then  the  two  polygons  are  similar,  §  331 

{two  polygons  composed  of  the  same  number  of  A  similar  to  each  other  and 
similarly  placed,  are  similar). 


mOBLEMS    OF    COMPUTATION. 


173 


Problems  of  Computation. 

219.   To  comi>ute  the  altitudes  of  a  triangle  in  terms  of  its  sides. 
C  C 


e   F    D 

Fig.  2. 


At  least  one  of  the  angles  ^  or  .S  is  acute.    Suppose  it  is  the  angle  B. 


In  the  A  CDB, 
In  the  A  ABC, 


Whence, 
Hence 


/t2  =  a2  -  BI)\ 

b^  =  a'  +  c^-2cxBD. 


338 
342 


BD 


4c2  ""  4c2 

;^  +  b^) 


__  (2ac  +  a^  +  c^-  b^){2ac  -  a' 

4c2 
^{(a  +  c)'-6'}{6'-(a-c)^} 

4c2 
_  (g  +  6  +  c)  (g  +  c  -  6)  (6  +  g  -  c)  (6  -  g  +  c) 
4c2 
g  +  6  +  c  =  2s. 
a  +  c  -  b  =  2{s  -  b), 
b  +  a  —  c  =  2{s  —  c), 
b  —  a  +  c  =  2{s  —  a). 
^2_  2s  X  2(s  -  g)  X  2(s  -  6)  X  2(s  -  c) 
4c2 

By  simplifying,  and  extracting  the  square  root, 
h  =  -  Vs'(s-g)(8-6)(s-c). 


Let 

Then 


Hence 


220.    To  compute  the  medians  of  a  triangle  in  terms  of  its  sides. 

By  I  344,  g2  +  62  =  2  m2  +  2  /'^  y.  (Fig.  2) 


Whence 


4m2  =  2(g2  +  6»)-c2. 
.-.  m  =  J  V2(g»  +  62)_c2. 


Uwyi::*^^-':)"   '^ 


174 


PLANE  GEOMETRY. 


BOOK    III. 


221.    To  coiupuLe  the  bisectors  of  a  triangle  in  terms  of  the  sides. 
By  g  349, 

By  'i  313. 


^2  = 

b 
.-.  AD  = 


ah-ADx  BD. 

BD__AD  +  BD 
a 
he 


a  +  6 


Whence 


t'  =-ah~ 


a  +  b 

and    BD 

abc'^ 


c_ 

a  +  b 

ac 

a  +  b 


4; 

^^ 

\T^ 

c 

D! 

A 

■'    '} 
J 

N^ 

/ 

^-.,^_ 

.ji'-''' 

Whence 


(a  +  by 

\        («  +  if) 
ab{{a  +  bf~c^} 
(a  +  bf 
—  <^^  (c^  +  ^  -f-  c)  (<^  +  ^  —  c) 
(a  +  bf 
abx2sx2(s-  c) 
{a  +  hf 

Vabs  {s^  —  c). 


o 

a+b 


222.    To  compute  the  radius  of  the  circle  circumscribed  about  a  tri- 
angle in  terms  of  the  sides  of  the  triangle.  ^ ^^   t 

By  §350,     ABxAO=-AExAD, 
or  bc  =  2Bx  AD. 

But  AD 


Whence 


Ji^ 


-  Vs  (s  —  a)  {s  —  6)  (s  —  c). 

a 

abc 


a)  (s  —  b){s  —  c) 

223.  If  the  sides  of  a  triangle  are  3,  4,  and  5,  is  the  angle  opposite  5 
right,  acute,  or  obtuse  ?   ^-^ 

224.  If  the  sides  of  a  triangle  are  7,  9,  and  12,  is  the  angle  opposite 
12  right,  acute,  or  obtuse  ?    '  ' 

225.  If  the  sides  of  a  triangle  are  7,  9,  and  11,  is  the  angle  opposite 
11  right,  acute,  or  obtuse  ?    ■ 

226.  The  legs  of  a  right  triangle  are  8  inches  and  12  inches;  find  the 
lengths  of  the  projections  of  these  legs  upon  the  hypotenuse,  and  the  dis- 
tance of  the  vertex  of  the  right  angle  from  the  hypotenuse. 

227.  If  the  sides  of  a  triangle  are  6  inches,  9  inches,  and  12  inches, 
find  the  lengths  (1)  of  the  altitudes  ;  (2)  of  the  medians  ;  (3)  oi  the  bisec- 
tors ;  (4)  of  the  radius  of  the  circumscribed  circle. 


EXERCISES.  175 


Theorems. 

^  228.  Any  two  altitudes  of  a  triangle  are  inversely  proportional  to 
the  corresponding  bases. 

V  229.  Two  circles  touch  at  P.  Through  P  three  lines  are  drawn,  meet- 
ing one  circle  in  A,  B,  C,  and  the  other  in  A^,  B\  C,  respectively.  Prove 
that  the  triangles  ABC,  A'B'C  are  similar. 

Si^SO.  Two  chords  AB,  CD  intersect  at  M,  and  A  is  the  middle  point  of 
the  arc  CD.  Prove  that  the  product  AB  x  -4i/ remains  the  same  if  the 
chord  AB  is  made  to  turn  about  the  fixed  point  A. 

Hint.  Draw  the  diameter  AE,  join  BE,  and  compare  the  triangles 
thus  fvyrmed. 

s  231.  The  sum  of  the  squares  of  the  segments  of  two  perpendicular 
chords  is  equal  to  the  square  of  the  diameter  of  the  circle. 

If  AB,  CD  are  the  chords,  draw  the  diameter  BE,  join  AC,  ED,  BD, 
and  prove  that  AC=^  ED.     Apply  g  338. 

^232.  In  a  parallelogram  ABCD,  a  line  DE  is  drawn,  meeting  the 
diagonal  AC  \n  F,  the  side  BC  in  G,  and  the  side  AB  produced  in  E. 
Prove  that  Df  =  FOx  FE. 

^233.   The  tangents  to  two  intersecting  circles  drawn  from  any  point    s/ 
in  their  common  chord  produced,  are  equal.     (§  348.)  "'^ 

^234.  The  common  chord  of  two  intersecting  circles,  if  produced,  will 
bisect  their  common  tangents.     (§  348.) 

■-  235,  If  two  circles  touch  each  other,  their  common  tangent  is  a  mean 
proportional  between  their  diameters. 

Hint.  Let  AB  be  the  common  tangent.  Draw  the  diameters  AC,  BD. 
Join  the  point  of  contact  P  to  A,  B,  C,  and  D.  Show  that  APD  and  BPC 
are  straight  lines  JL  to  each  other,  and  compare  A  ABC,  ABD. 

236.  If  three  circles  intersect  one  another,  the  common  chords  all  pass 
through  the  same  point. 

Hint.  Let  two  of  the  chords  AB  and  CD 
meet  at  0.  Join  the  point  of  intersection  E 
to  G,  and  suppose  that  EO  produced  meets 
the  same  two  circles  at  two  different  points  P 
and  Q.  Then  prove  that  0P=  OQ;  hence, 
\     that  the  points  P  and  Q  coincide. 


176 


PLANE   aEOMETRY. — BOOK    III. 


""237.  If  two  circles  are  tangent  internally,  all  chords  of  the  greater 
circle  drawn  from  the  point  of  contact  are  divided  proportionally  by  the 
circumference  of  the  smaller  circle. 

Hint.  Draw  any  two  of  the  chords,  join  the  points  where  they  meet 
the  circumferences,  and  prove  that  the  A  thus  formed  are  similar. 

^238.  In  an  inscribed  quadrilateral,  the  product  of  the  diagonals  is 
equal  to  the  sum  of  the  products  of  the  opposite  sides. 

nmT.   Dr&vf  DJS,  making  ZCDU=ZADB.    The      ^^^ 
A  ABD  and  CDE  are  similar.     Also  the  A  BCD  and 
ADE  are  similar. 

239.  The  sum  of  the  squares  of  the  four  sides  of 
any  quadrilateral  is  equal  to  the  sum  of  the  squares 
of  the  diagonals,  increased  by  four  times  the  square 
of  the  line  joining  the  middle  points  of  the  diagonals. 

Hint.  Join  the  middle  points  F,  E,  of  the  diag- 
onals. Draw  EB  and  ED.  Apply  ^  344  to  the 
A  ABC  and  ADC,  add  the  results,  and  eliminate 
BE^  +  DE''  by  applying  g  343  to  the  A  BDE. 

'"  240.   The  square  of  the  bisector  of  an  exterior  angle  of  a  triangle  is 
equal  to  the  product  of  the  external  segments  deter- 
mined by  the  bisector  upon  one  of  the  sides, dimin- 
ished by  the  product  of  the  other  two  sides. 

Hint.   Let  CD  bisect  the  exterior  Z  BCH  of 
the  A  ABC.    Circumscribe  a  O  about  the  A,  pro- 
duce DC  to  meet  the  circumference  in  F,  and  draw  BF. 
BCF  similar.     Apply  §  347. 

"  241.  If  a  point  0  is  joined  to  the  vertices  of  a  triangle  ABC,  and 
through  any  point  A^  in  OA  a  line  parallel  to  AB  is  drawn,  meeting  OB 
at  B^,  and  then  through  B^  a  line  parallel  to  BC,  meeting  OC  at  C^, 
and  (7  is  joined  to  A^,  the  triangle  A^B'C  will  be  similar  to  the  tri- 
angle ABC 

242.  If  the  line  of  centres  of  two  circles  meets  the  circumferences  at 
the  points  A,  B,  C,  D,  and  meets  the  common  exterior  tangent  at  P,  then 
FAxFD  =  PBx  FC 

"^  243.  The  line  of  centres  of  two  circles  meets  the  common  exterior 
tangent  at  P,  and  a  secant  is  drawn  from  P,  cutting  the  circles  at  the 
consecutive  points  E,  F,  G,  H.     Prove  that  FExPH^-  PFx  PG. 


Trove  ^ACD, 


■7i 


ft    EXEBCISES.  .  177 

L 

Numerical  Exercises.      ^  -  i'- 

"^244".  A  line  is  drawn  parallel  to  a  side  AB  of  a  triangle  ABC,  and 
cutting  *AC  in  D,  BC  in  E.  U  AD.  DC  =2:  3,  and  AB  =  20  inches, 
find  DE.     l^j  ^   ^     J  - 

^245.   The  sides  of  a  triangle  are  9,  12,  15.    Find  the  segments  made  by'  '^'' 
bisecting  the  angles.    (^  313.)  "     .     ^^ 

^^  246.   A  tree  casts  a  shadow  90  feet  long,  when  a  vertical  rod  6  fe©^  *^'^*'. 
high  casts  a  shadow  4  feet  long.     How  high  is  the  tree  ?/'!-' 

^  247.  The  bases  of  a  trapezoid  are  represented  by  a,  b,  and  the  altitude 
by  h.  Find  the  altitudes  of  the  two  triangles  formed  by  producing  the 
legs  till  they  meet. 

"^  248.  The  sides  of  a  triangle  are  6,  7,  8.  In  a  similar  triangle  the  side 
homologous  to  8  is  equal  to  40.     Find  the  other  two  sides.    **!.''''     _     - 

'  249.  The  perimeters  of  two  similar  polygons  are  200  feet  and  300  feet. 
If  a  side  of  the  first  polygon  is  24  feet,  find  the  homologous  side  of  the 
second  polygon,    3  w 

250.  How  long  must  a  ladder  be  to  reach  a  window  24  feet  high,  if 
the  lower  end  of  the  ladder  is  10  feet  from  the  side  of  the  house  ?  '^  w  *-^ 

\^251.   If  the  side  of  an  equilateral  triangle  =  a,  find  the  altitude,    jf'7' 

^  2o2.   If  the  altitude  of  an  equilateral  triangle  =  h,  find  the  side.  -    " 

^  253.  Find  the  lengths  of  the  longest  and  the  shortest  chord  that  can 
be  drawn  through  a  point  6  inches  from  the  centre  of  a  circle  whose 
radius  is  equal  to  10  inches.     T  /  C> 

^  254.   The  distance  from  the  centre  of  a  circle  to  a  chord  10  inches  long     r- 
is  12  inches.    Find  the  distance  from  the  centre  to  a  chord  24  inches  long.     ^ 

"  255.  The  radius  of  a  circle  is  5  inches.  Through  a  point  3  inches  from 
the  centre  a  diameter  is  drawn,  and  also  a  chord  perpendicular  to  the 
diameter.  Find  the  length  of  this  clil^^,  and  the  distance  from  one  end 
of  the  chord  to  the  ends  of  the  diametter.     'V  —     *■       '^/;  -^ 

'^56.   The  radius  of  a  circle  is  6  inches.     Through  a  point  10  inches 
from  the  centre  tangents  are  drawn.     Find  the  lengths  of  the  tangents,  Vi^6 
and  also  of  the  chord  joining  the  points  of  contact.     XO  •  ^  - 

257.  If  a  chord  8  inches  long  is  3  inches  distant  from  the  centre  of 
the  circle,  find  the  radiuinand  the  distances  from  the  end  of  the  chord  to 
the  ends  of  the  diameter  which  bisects  the  chord.    ■Sf' ^     •'        //,  '-/-^ 


178  PLANE    GEOMETRY.  —  BOOK    III. 

258.  The  radius  of  a  circle  is  13  inches.  Through  a  point  5  inches 
from  the  centre  any  chord  is  drawn.  What  is  the  product  of  the  two  seg- 
mer^^^ii%^e  chord  ?  What  is  the  length  of  th)j  sljoftest  chord  that  can 
be  drawn  through  the  point  ?  '        / 

259.  From  the  end  of  a  tangent  20  inches  long  a  secant  is  drawn 
through  the  centre  of  the  circle.  If  the  exterior  segment  of  this  secant 
is  8  inches,  find  the  radius  of  the  circle.  "^J^  / 

— ^260.  The  radius  of  a  circle  is  9  inches  ;  the  length  of  a  tangent  is  12 
inches.  Find  the  length  of  a  secant  drawn  from  the  extremity  of  the 
tangent  to  the  centre  of  the  circle. 

261.  The  radii  of  two  circles  are  8  inches  and  3  inches,  and  the  dis- 
tance between  their  centres  is  15  inches.  Find  the  lengths  of  their  com- 
mon tangents.     /  - '  •^' 

"^  262.    Find  the  segments  of  a  line  10  inches  long  divided  in  extreme 
and  mean  ratio. 

263.  The  sides  of  a  triangle  are  4,  5,  6.  Is  the  largest  angle  acute, 
right,  or  obtuse  ?  t^- 

Problems. 

>--  264.  To  divide  one  side  of  a  given  triangle  into  segments  proportional 
to  the  adjacent  sides.     (§  313.) 

^    265.   To  produce  a  line  AB  to  a  point  Cso  that  AB  -.  AC=3:5. 

'"  266.  To  find  in  one  side  of  a  given  triangle  a  point  whose  distances 
from  the  other  sides  shall  be  to  each  other  in  a  given  ratio. 

267.  Given  an  obtuse  triangle ;  to  draw  a  line  from  the  vertex  of  the 
obtuse  angle  to  the  opposite  side  which  shall  be  a  mean  proportional 
between  the  segments  of  that  side. 

■^  268.  Through  a  given  point  P  within  a  given  circle  to  draw  a  chord 
AB  so  that  AP:  BP=  2:3. 

269.  To  draw  through  a  given  point  P  in  the  arc  subtended  by  a  chord 
AB  2u  chord  which  shall  be  bisected  by  AB. 

270.  To  draw  through  a  point  P,  exterior  to  a  given  circle,  a  secant 
PAB  so  that  PA :  ^P  =  4  :  3. 

271.  To  draw  through  a  point  P,  exterior  to  a  given  circle,  a  secant 
PAB  so  that  A^  ^PAx  PB. 

272.  To  find  a  point  Pin  the  arc  subtended  by  a  given  chord  AB  so 
that  PA:PB  =  3:  1. 


EXERCISES.  179 

■273.  To  draw  through  one  of  the  points  of  intersection  of  two  circles 
a  secant  so  that  the  two  chords  that  are  formed  shall  be  to  each  other 
in  the  ratio  of  3  :  5. 

^274,   To  divide  a  line  into  three  parts  proportional  to  2,  |,  ^. 

v275.  Having  given  the  greater  segment  of  a  line  divided  in  extreme 
and  mean  ratio,  to  construct  the  line. 

~276.  To  construct  a  circle  which  shall  pass  through  two  given  points 
and  touch  a  given  straight  line. 

277.  To  construct  a  circle  which  shall  pass  through  a  given  point  and 
touch  two  given  straight  lines. 

278.  To  inscribe  a  square  in  a  semicircle. 

279.  To  inscribe  a  square  in  a  given  triangle. 

Hint.   Suppose  the  problem  solved,  and  DEFO  the  inscribed  square. 

Draw  Clf  II  to -4-5,  and  let  j4i^  produced  meet  •      r^                     -^^ 
CM'm  M.  Draw  Off  and  MN  Jl  to  AB,  and 
produce  A  Bio  meet  MN  at  N.   The  A  ACM, 

AOF  are  similar;    also  the  A  AMN,  AFE  Ql 

are  similar.     By  these   triangles   show  that  /l 

the  figure  CMNH  is  a  square.   By  construct-  LA 

ing  this-square,  the  point  F  can  be  found.  A  L 

280.  To  inscribe  in  a  given  triangle  a  rectangle  similar  to  a  given 
rectangle. 

281.  To  inscribe  in  a  circle  a  triangle  similar  to  a  given  triangle. 

282.  To  inscribe  in  a  given  semicircle  a  rectangle  similar  to  a  given 
rectangle. 

283.  To  circumscribe  about  a  circle  a  triangle  similar  to  a  given 
triangle. 


284.   To  construct  the  expression,  x  -  ^^-^ ;  that  is,  —  x 


c 


de  d        e 

285.  To  construct  two  straight  lines,  having  given  their  sum  and 
their  ratio. 

286.  To  construct  two  straight  lines,  having  given  their  difference 
and  their  ratio. 

287.  Having  given  two  circles,  with  centres  0  and  0\  and  a  poiiit  A 
in  their  plane,  to  draw  through  the  point  A  a  straight  line,  meeting'the 
<:ircumferences  at  B  and  C,  so  that  AB  -.  AO^  1:2. 

Hint.  Suppose  the  problem  solved,  join  OA  and  produce  it  to  D. 
making  OA.  AD^}  -.2.     Join  DC;  ^  OAB.  A  DC  are  similar. 


BOOK   IV. 
AREAS    OF    POLYGONS. 

358,  The  area  of  a  surface  is  the  numerical  measure  of  the 
surface  referred  to  the  unit  bf  surface. 

The  unit  of  surface  is  a  square  whose  side  is  a  unit  of  length ; 
as  the  square  inch,  the  square  foot,  etc. 

359.  Equivalent  figures  are  figures  having  equal  areas. 


Proposition  I.     Theorem. 

360.  The  areas  of  two  rectangles  having  equal  alti- 
tudes are  to  each  other  as  their  bases. 

Di : ! ! ' 1 ^ ^c  n 


B 


0  "       -       o  /^ 

Let  the  two  rectangles  be  AC  and  AF,  having  the 

same  altitude  AD. 

rrj  red.  AC      AB 

10  prove  -— 1  =  — -— • 

^  rect.^i^     AE 

Proof.   Case  I.    When  AB  and  AE  are  commensurable. 

Suppose  AB  and  AE  have  a  common  measure,  as  AO, 

which  is  contained  in  AB  seven  times  and  in  AE  four  times. 

Then  if  =  4-  '         W 

Apply  this  measure  to  AB  and  A  ll\  and  at  the  several 
points  of  division  erect  Js. 

The     rect.  ^C'will  be  divided  into  seven  rectangles, 
and  the  rect.  ^jPwill  be  divided  into  four  rectangles. 


AREAS   OF    POLYGONS. 

These  rectangles  are  all  equal. 

rect.  AC      7 


Hence 

From  (1)  and  (2) 


rect.^i^      4 
rect.  ^6^  _^^ 
rect.^i^      AU 

Case  II.    When  AB  and  AE  are  incommensurable. 
D> ^  f^- 


181 

§186 
(2) 

Ax.  1 


B 


D 


K 


E' 


Divide  AB  into  any  number  of  equal  parts,  and  apply  one 
of  them  to  AE  as  often  as  it  will  be  contained  in  AE. 

Since  AB  and  AE  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  from  ^4  to  a  point  K,  leaving  a 
remainder  KE  less  than  one  of  the  parts. 
Draw  KH II  to  EF. 
Since  AB  and  ^iTare  commensurable, 
xQGi.AH^AK 

ab' 


Case  I. 


rect.  A  C 

These  ratios  continue  equal,  as  the  unit  of  measure  is  indefi- 
nitely diminished,  and  approach  indefinitely  the  limiting  ratios 


rect.^i^       .  AE  ,.     , 

— T  and     -^  respectively. 

rect.^C  AB      ^  ^ 

.  rect.  ^i^ 


AE 

ab' 


§250 


rect.  AC 

{if  two  variables  are  constantly  equal,  and  each  approaches  a  limit,  the 
limits  are  equal).  q  ^  q 

361.  Cor.  The  areas  of  two  rectangles  having  equal  bases  are 
to  each  other  as  their  altitudes.  For  AB  and  AE  may  be  con- 
sidered as  the  altitudes,  AD  smd  AB  b.s  the  bases. 

Note.  In  propositions  relating  to  areas,  the  words  "rectangle," 
"triangle,"  etc.,  are  often  used  for  "area  of  rectangle,"  "area  of  tri- 
angle," etc. 


182 


PLANE    GEOMETRY. BOOK    IV. 


Proposition  II.     Theorem. 

362.   The  areas  of  two  rectangles  are  to  each  other 
as  the  products  of  their  bases  hy  their  altitudes. 


R 


h  h'  b 

Let  11  and  B'  be   two  rectangles,   having  for  their 
bases  h  and  h',  and  for  their  altitudes  a  and  a'-. 
R  _  aX  b  ■ 
El      a'  X  h^' 

Proof.    Construct  the  rectangle  8,  with  its  base  the  same  as 
that  of  R,  and  its  altitude  the  same  as  that  of  R\ 

R      a 


To  prove 


Then 


8 


§361 


and 


{^rectangles  having  equal  bases  are  to  each  other  as  their  altitudes) 

8_^h 
R'     V 

{rectangles  having  equal  altitudes  are  to  each  oilier  as  their  bases) 
By  multiplying  these  two  equalities, 

R  __  aXb 

R'      a'xb''  Q.i 


§  360 


Ex.  288.  Find  the  ratio  of  a  rectangular  lawn  72  yards  by  49  yards 
to  a  grass  turf  18  inches  by  14  inches. 

Ex.  289.  Find  the  ratio  of  a  rectangular  courtyard  18^  yards  by  ]5h 
yards  to  a  flagstone  31  inches  by  18  inches. 

Ex.  290.  A  square  and  a  rectangle  have  the  same  perimeter,  100  yards. 
The  length  of  the  rectangle  is  4  times  its  breadth.     Compare  their  areas. 

Ex.  291.  On  a  certain  map  the  linear  scale  is  1  inch  to  5  miles.  How 
many  acres  are  represented  on  this  map  by  a  square  the  perimeter  of 
whirli  is  1  inch  ? 


AREAS   OF    ruLYGoNS. 


183 


Proposition  III.     Theorem. 

363.   The  area  of  a  rectangle  is  equal  to  the  product 
of  its  base  and  altitude. 


s 


Let  R  be  the  rectangle,  b  the  base,  and  d  the  alti- 
tude; and  let  U  be  a  square  whose  side  is  equal  to 
the  linear  unit. 

To  prove  the  area  of  R  ---  axh.  • 

R  ^axh 
'U     1x1 

{ixioo  rectan(/les  are  to  each  other  as  the  product  of  their  bases  and  altitudes). 


=  axb, 


§362 


But 


—  =  the  area  of  R. 


§358 


.*.  the  area  of  R  =  a  X  b.  q.  e.  d. 

364.  Scholium.  When  the  base  and  altitude  each  contain 
che  linear  unit  an  integral  number  of  times,  this  proposition  is 
rendered  evident  by  dividing  the  figure  into   squares,  each 


equal  to  the  unit  of  measure.  Thus,  if  the  base  contain  seven 
linear  units,  and  the  altitude  four,  the  figure  may  be  divided 
into  twenty-eight  squares,  each  equal  to  the  unit  of  measure  ; 
and  the  area  of  the  figure  equals  7x4  units  of  surface. 


184 


PLANE   GEOMETRY.  —  BOOK   IV. 


Proposition  IV.     Theorem. 

366.  The  area  of  a  parallelogram  is  equal  to  the 
product  of  its  base  and  altitude. 
BE  C     F 


A  h  D  A         h         D 

Let  AEFD  be  a  parallelogram,  AD  its  base,  and  CD 
its  altitude. 

To  prove  the  area  of  the  O  AEFB  =--  AD  X  CB. 
Proof.    From  A  draw  AB  II  to  BCio  meet  i^.^ produced. 
Then  the  figure  ABCB  Avill  be  a  rectangle,  with  the  same 
base  and  altitude  as  the  O  AEFB. 
In  the  rr.  A  ABE  and  BCF 

AB-^CB  and  AE^  BE,  §  179 

{heing  opposite  sides  of  a  CJ). 

.'.AABE^ABCF,  §161 

{two  rt.  ^  are  equal  when  the  hypotenuse  and  a  side  of  the  one  are  equal 
respectively  to  the  hypotenuse  and  a  side  of  the  other). 

Take  away  the  A  BCF,  and  we  have  left  the  rect.  ABCB. 

Take  away  the  A  ABE,  and  w^  have  left  the  O  AEFB. 

.-.  rect.  ABCB  o  O  AEFB.  Ax.  3 

But  the  area  of  the  rect.  ABCB  =  axb,         §  363 

.-.  the  area  of  the  O  AEFB  =  axb.  Ax.  1 

Q.  E.  O. 

366.  Cor.  1.  Parallelogravis  having  equal  bases  and  equal 
altitudes  are  equivalent. 

367.  Cor.  2.  Parallelograms  having  equal  bases  are  to  each 
other  as  their  altitudes ;  parallelograms  having  equal  altitudes 
are  to  each  other  as  their  bases ;  any  two  parallelograms  are 
to  each  other  as  the  products  of  their  bases  by  their  altitudes. 


AREAS   OF    POLYGONS. 


185 


Proposition  V.     Theorem. 

368.  The  area  of  a  triangle  is  equal  to  one-half  of 
the  product  of  its  base  by  its  altitude. 


§168 


§178 


B      D 


Let  ABC  be  a  triangle,  AB  its  base,  and  DC  its 
altitude. 

To  prove  the  area  of  the  A  ABC— ^  AB  X  DC. 
Proof.  ,      From  C  draw  CH  II  to  AB. 

From  A  draw  J  ^  II  to  BC. 

The  figure  ABCH'i^  a  parallelogram, 
{having  its  opposite  sides  parallel), 

and  AC ia  its  diagonal. 

.■.AABC=AAJa:C, 
(the  diayonal  of  a  CD  divides  it  into  two  ec^ual  ^). 

The  area  of  the  O  ABCJET  is  equal  to  the  product  of  its 
base  by  its  altitude.  §  365 

Therefore  the  area  of  one-half  the  CJ,  that  is,  the  area  of 
the  A  ABC,  is  equal  to  one-half  the  product  of  its  base  by  its 
altitude. 

Hence,     the  area  of  the  A  ABG=\AB  X  DC. 

Q.  E.  D. 

369.  Cor.  1.  Triangles  having  equal  bases  and  equal  alti- 
tudes are  equivalent. 

370.  Cor.  2.  Tnangles  having  equal  bases  are  to  each  other 
as  their  altitiides ;  triangles  having  equal  altitudes  are  to  each 
other  as  their  bases ;  any  two  triangles  are  to  each  other  as  the 
product  of  their  hoses  by  their  altitudes. 


186 


PLANE    GEOMKTEY. 


BOOK    IV. 


Proposition  VI.     Theorem. 

371.  The  area  of  a  trapezoid  is  equal  to  one-half 
the  sum  of  the  parallel  sides  multiplied  hy  the  alti- 
tude. 


H             E     h' 

C 

qL 

1 

"\ 

\r 

a 

\ 

A.  FOB 

Let  ABCH  be  a  trapezoid,  and  EF  the  altitude. 

To  prove         area  of  ABCH  =  i  {HC-\-  AB)  EF. 
Proof.  Draw  the  diagonal  AG. 

Then  the  area  of  the  A  ABC=  \  (AB  X  UF),  §  368 

and        the  area  of  the  A  AirO=  I  {HCx  EF). 

By  adding,     area  of  ABCH=  J  {AB  +  HC)  EF.       q.  e.  d. 

372.  Cor.  The  area  of  a  trapezoid  is  equal  to  the  product 
of  the  tnedian  hy  the  altitude.  For,  by  §  191,  OF  is  equal  to 
\{HC-\-AB)\  and  hence 

the  area  of  ABCH=  OP  X  EF. 

373.  Scholium.  The  area  of  an  irregular  polygon  may  be 
found  by  dividing  the  poly- 
gon into  triangles,  and  by 
finding  the  area  of  each  of 
these  triangles  separately. 
But  the  method  generally 
employed  in  practice  is  to 
draw  the.  longest  diagonal, 
and  to  let  fall  perpendiculars  upon  this  diagonal  from  the 
other  angular  points  of  the  polygon. 

The  polygon  is  thus  divided  into  right  triangles  and  trape- 
zoids ;  the  sum  of  the  areas  of  these  figures  will  be  the  area 
of  the  polygon. 


L 


AREAS   OF    POLYGONS. 


18; 


Proposition  VII.     Theorem. 

374.  The  areas  of  two  tidangles  which  have  an  angle 
of  the  one  equal  to  an. angle  of  the  other  are  to  each 
other  as  the  products  of  the  sides  including  tlxe  equal 
angles. 


Let  the  triangles  ABC  and  ADE  have  the  common 

angle  A. 

n.  A  ABC      ABxAC 

10  prove 

Proof. 
Now 


and 


A  ADE     AD  X  AE 

Draw  BE. 

A  ABC 

AC 

A  ABE 

AE 

A  ABE_ 

_AB 

A  ADE     AD 

{^  having  the  same  altitude  are  to  each  other  as  their  bases). 

By  multiplying  these  equalities, 

AABC_ABxAC 
A  ADE     AD  X  AE 


§370 


Q.  E.  D. 


Ex.  292.  The  areas  of  two  triangles  which  have  an  angle  of  the  one 
supplementary  to  an  angle  of  the  other  are  to  each  other  as  the  products 
01  the  sides  including  the  supplementary  angles. 


188 


PLANE   GEOMETEY,  —  BOOK    IV. 


Comparison  of  Polygons. 

Proposition  VIII.     Theorem. 

375.  The  areas  of  two  similar  triangles  are  to  each 
other  as  the  squares  of  any  two  homologous  sides. 


A  o  D  A'    O'  Ji' 

Let  the  two  triangles  be  ACB  and  A'C  B'. 
AAOB        AB" 


To  prove 


A  A^C^B^      JTb'' 


Draw  the  perpendiculars  CO  and  CO'. 

A  ACB         AB  X  CO        AB 


Then 


CO 


A  A'C'B'     A'B'  X  Ca     A'B'  ^  C'O'^ 


§370 


{two  A  are  to  each  other  as  the  products  of  their  bases  by  their  altitudes). 

But  ^=^:  §328 

(the  homologous  altitudes  of  similar  A  have  the  same  ratio  as  their  homolo- 
gous bases). 

CO    •  AB 

Substitute,  in  the  above  equality,  for     ^      its  equal 


then 


CO' 
AB       AB" 


A  ACB        AB 


A  A'C'B'      A'B'     A'B'      jJB' 


A'B'' 


Q.  E.  D 


COMPARISON  OF  POLYGONS. 


189 


Pboposition  IX.     Theorem. 

376.  The  areas  of  two  similar  polygons  are  to  each 
other  as  tJie  squares  of  any  two  homologous  sides. 


ir         c' 

Let  S  and  S^  denote  the  areas  of  the  two  similar 
polygons  ABC,  etc.,  and  A'B'C,  etc. 

To  prove  8:8'  =  IF  :  IB^. 

Proof.  By  drawing  all  the  diagonals  from  the  homologous 
\rertices  E  and  E\  the  two  similar  polygons  are  divided  into 
triangles  similar  and  similarly  placed.  §  332 


AB' 


A  ABE 


A'B' 


AA'B'E' 

Kc'E'V 


rBE!_\ 
KWe'V 

A  CBE 


A  BCE 
A  B'C'E' 


AC'B'E'  ^    ' 

{similar  A  are  to  each  other  as  the  squares  of  any  two  homologous  sides). 
■      ,       .  A  ABE  _  A  BCE  _^    A  CBE 

inatis,      '^j^,^,^,     '^B'C'E     AC'jD'E'' 

.        A  ABE-\-  BCE+  CBE       __  A  ABE  ^  AB^     .  oqq 
"  AA'B'E' -{-B'C'E'-^C'B'E'      AA'B'E'      A'W 
{in  a  series  of  equal  ratios  the  sum  of  the  antecedents  is  to  the  sum  of  the 
consequents  as  any  antecedent  is  to  its  consequent). 


8:8'  =  AB":A'B' 


Q.  E.  O. 


377.  Cor.  1.   The  areas  of  two  similar  polygons  are  to  each 
of  he)'  as  the  squares  of  any  two  homologous  lines. 

378.  Cor.  2.   The  homologous  sides  of  two  similar  polygons 
have  the  same  ratio  as  the  square  roots  of  their  areas. 


190 


PLANE    GEOMETRY. 


BOOK    IV. 


Proposition  X.    ^Theorem. 

379.  The  square  described  on  the  hypotenuse  of  a 
right  triangle  is  equivalent  to  the  sum  of  the  squares 
on  the  other  two  sides.,  ' 

Let  BE,  GH,  AF,  be  squares  on  the  three  sides  of  the 
right  triangle  ABC. 

To  prove  BC'  =0=  AB'  +  AC\ 

Proof.  Through  A  draw  ^X  II  to 
CE,  and  draw  AB  and  FC. 

Since  A  BAG,  BAG,  and  CAII 
are  rt.  A,  CAG  and  BAIT  are 
straight  lines. 

Since  BB  ~  BC,  being  sides  of 
the  same  square,  and  BA  —  BT] 
for  the  same  reason,  and  since 
Z  ABB  =  ZFBC,  each  being  the 
sum  of  a  rt.  Z  and  the  Z  ABC, 

the  A  ABB  =  A  FBC. 

Now       the  rectangle  BB  is  double  the  A  ABB, 
[having  the  same  base  BD,  and  the  same  altitude,  the  distance  between  the 
lis  AL  and  BD), 

and  the  square^i^is  double  the  A  FBC, 

{having  the  same  base  FB,  and  the  same  altitude,  the  distance  between  the 
lis  FB  and  OC). 

Hence  the  rectangle  BB  is  equivalent  to  the  square  AF. 

In  like  manner,  by  joining  ^^and  BK,  it  may  be  proved 
that  the  rectangle  CB  is  equivalent  to  the  square  CH. 

Therefore  the  square  BE,  which  is  the  sum  of  the  rectangles 
BL  and  CB,  is  equivalent  to  the  sum  of  the  squares  CH  and 

^F.  .  Q.  E.  D. 

380.  Cor.  The  square  on  either  leg  of  a  right  triangle  is 
equivalent  to  the  difference  of  the  squares  on  the  hypotenuse  and 
the  other  leg,  X 


§15U 


COMPARISON    OF    POLYGONS. 


191 


Ex.  293,  The  square  constructed  upon  the  sum  of  two  straight  lines 
is  equivalent  to  the  sum  of  the  squares  constructed  upon  these  two  line£, 
increased  by  twice  the  rectangle  of  these  lines. 

Let  AB  and  BC  be  the  two  straight  lines,  and  AC  their  sum.  Con- 
struct the  squares  ACOK  and  ABED  upon  ^Cand 
AB  respectively.  Prolong  BE  and  DE  until  they 
meet  KG  and  CO  respectively.  Then  we  have  the 
square  EFGH,  with  sides  each  equal  to  BC.  Hence, 
the  square  ACOK  is  the  sum  of  the  squares  ABED  J) 
and  EFOH,  and  the  rectangles  DEHK  and  BCFE, 
the  dimensions  of  which  are  equal  to  AB  and  BC. 


A 


K 


BC 


E 

H    O 


Ex.  294.  The  square  constructed  upon  the  difference  of  two  straight 
lines  is  equivalent  to  the  sum  of  the  squares  constructed  upon  these  two 
lines,  diminished  by  twice  the  rectangle  of  these  lines. 

Let  AB  a,nd  AC  he  the  two  straight  lines,  and  BC  their  difference. 


L 


C 

D 

B 


E 


Construct  the  square  ABFG  upon  AB,  the  square      jj^     j^ 
ACKH Vi^on  AC,  and  the  square  BEDC  upon  BCios 
shown  in  the  figure).     Prolong  ED  until  it  meets  AG 
in  L. 

The  dimensions  of  the  rectangles  LEFG  and  HKDL 
are  AB  and  AC,  and  the  square  BCDE  is  evidently 
the  difference  between  the  whole  figure  and  the  sum 
of  these  rectangles ;   that  is,  the   square   constructed      (^  '^ 

upon  BC  is  equivalent  to  the  sum  of  the  squares  constructed  upon  AB 
and  AC  diminished  by  twice  the  rectangle  of  AB  and  AC. 

Ex.  295.  The  difference  between  the  squares  constructed  upon  two 
straight  lines  is  equivalent  to  the  rectangle  of  the  sum  and  difference  of 
these  lines. 

Let  ABDE  and  BCGF  be  the  squares  constructed  upon  the  two 
straight  lines  AB  and  BC.  The  difference  between 
these  squares  is  the  polygon  ACOFDE,  which  poly- 
gon, by  prolonging  CO  to  H,  is  seen  to  be  composed  of 
the  rectangles  ACHE  and  OFDH.  Prolong  AE  and 
CHto  J  and  A"  respectively,  making  j^JJand  I£K  each 
equal  to  BC,  and  draw  IK.  The  rectangles  OFDH 
and  EHKI  are  equal.  The  difference  between  the 
squares  ABDE  and  BCOF  is  then  equivalent  to  the 
rectangle  ACKI,  which  has  for  dimensions  AI=  AB  +  BC,  and  EH 
=  AB-BC 


K 


E 


H 

a 

D 


A 


C     B 


192 


PLANE    GEOMETRY, 


BOOK    IV. 


Problems  of  Construction.     , 

Proposition  XI.     Problem. 

381.  To  construct  a  square  equivalent  to  the  sum 
of  two  given  squares. 


Bl 


A^ 


Let  R  and  B'  be  two  given  squares. 

To  construct  a  square  equivalent  to  R'  +  Ti. 
Construction.  Construct  the  rt.  Z  A. 

Take  A  G  equal  to  a  side  of  H', 

AB  equal  to  a  side  of  R\  and  draw  BC. 

Construct  the  square  S,  having  each  of  its  sides  equal  to  BC. 

S  is  the  square  required. 

Proof.  BO'  -  AC'  +  AB\  §  379 

{the  square  on  the  hypotenuse  of  a  rt.  A  is  equivalent  to  the  sum  of  the 
squares  on  the  two  sides). 


.'.S-B'  +  R. 


Q.  E.  F. 


Ex.  296.  If  the  perimeter  of  a  rectangle  is  72  feet,  and  the  length  is 
equal  to  twice  the  width,  find  the  area. 

Ex.  297.  How  many  tiles  9  inches  long  and  4  inches  wide  will  be 
required  to  pave  a  path  8  feet  wide  surrounding  a  rectangular  court  120 
feet  long  and  36  feet  wide  ? 

Ex.  298.  The  bases  of  a  trapezoid  are  16  feet  and  10  feet ;  each  leg 
is  equal  to  5  feet.     Find  the  area  of  the  trapezoid. 


PROBLEMS   OF   CONSTRUCTION.  193 


Proposition  XII.     Problem. 


382.  To  construct  a  square  equivalent  to  the  differ- 
ence of  two  given  squares. 

I  \  /  \    A 


R 


Let  B  be  the  smaller  square  and  B'  the  larger. 

To  construct  a  square  equivalent  to  PJ  —  R. 
Oonstmction.  Construct  the  rt.  Z  A. 

Take  AB  equal  to  a  side  of  R. 

From  ^  as  a  centre,  with  a  radius  equal  to  a  side  of  R\ 

describe  an  arc  cutting  the  line  ^Xat  C. 

Construct  the  square  8,  having  each  of  its  sides  equal  to  A  0. 

8  is  the  square  required. 

Proof.  AC^  =0=  J^  _  aW;  §  380 

{the  square  on  either  leg  of  a  rt.  A  is  equivalent  to  the  difference  of  the 
squares  on  the  hypotenuse  and  the  other  leg). 

.-.  8^  R'  -  R. 

Q.  E.  F. 

Ex.  299.  Construct  a  square  equivalent  to  the  sum  of  two  squares 
whose  sides  are  3  inches  and  4  inches. 

Ex.  300.  Construct  a  square  equivalent  to  the  difference  of  two 
squares  whose  sides  are  2^  inches  and  2  inches. 

Ex.  30L  Find  the  side  of  a  square  equivalent  to  the  sum  of  two 
squares  whose  sides  are  24  feet  and  32  feet. 

Ex.  302.  Find  the  side  of  a  square  equivalent  to  the  difference  of  two 
squares  whose  sides  are  24  feet  and  40  feet. 

Ex.  303.  A  rhombus  contains  100  square  feet,  and  the  length  of  on( 
diagonal  is  10  feet.     Find  the  length  of  the  other  diagonal. 


194  PLANE    GEOMETRY. BOOK    IV. 


Proposition  XIII.     Problem. 

383.   To  construct  a  square,  equivalent  to  the  sum 
of  any  number  of  given  squares. 

II 

y    \ 

\ 

/  \ 

/     \ 

n /    -V...^         \       \ 

o •  /  ■■■--..        \      \ 


r 


::i^; 


Let  m,  n,  o,  p,  r  be  sides  of  the  given  squares. 

To  construct  a  square  o=  m'^  +  n^  +  o^  -i-p^  +  r\ 
Oonstruction.  Take  AB  =  m. 

Draw  AC  =n  and  ±  to  AB  at  A,  and  draw  BC. 

Draw  CU  =  a  and  _L  to  BC  at  C,  and  draw  BU. 

Draw  TJF  =p  and  ±  to  BE  at  E,  and  draw  BF. 

Draw  FII=  r  and  ±  to  BE  at  E,  and  draw  BH. 
The  square  constructed  on  BITis,  the  square  required. 


Proof.   BE'- -  EH  -^  BE\ 

-fh'-^ef'  +  eb'} 

-  EH"  +  EF''  -+(:^'  +  C^} 

<^  EH"  -{-  EG!-\-  EF  -\^CAl  +  aS\     §  379 

(<Ae  ium  of  the  squares  on  the  two  legs  of  a  rt.  A  is  equivalent  to  the  square 
on  the  hypotenuse). 

That  is,  BIT'  ^  m'  +  n'  +  o'  -|-  p'  +  r^ 

a  E.  F. 


PROBLEMS   OF   CONSTRUCTION. 


195 


Proposition  XIV.     Problem. 

384.  To  construct  a  polygon  similar  to  two  given 
similar  polygons  and  equivalent  to  their  sum. 


R' 


Os 


Bf  A"  Li"  P  11 

Let  12  and  I"  be  two  similar  polygons,  and  AB  and 
A'B'  two  homologous  sides. 

To  consto^ucitjt  similar  polygon  equivalent  to  R-\-  R\ 

Construction.  Construct  the  rt.  Z  P. 

Take  PJ7=  A'B,  cand  PO  =  AB. 

Draw  OH,  and  take  ^"^"  -  OIL 
Upon  A"B",  homologous  to  AP,  construct  ^"  similar  to  P. 
Then  P"  is  the  polygon  required. 

Proof.    PO'  +  PJI"  -  OlF\  .-.  AP'  4-  A^'"  =  A^Wl 
R        AP' 


Now 


and 


4"      ]|"i>^"' 


376 


{similar  polygons  are  to  each  other  as  the  squares  of  their  homologous  sides). 

P^B_AP^  +  A^P^ 
P'' 


By  addition, 


1. 


A"P"' 


Q.  E.  F 


196  PLANE   GEOMETRY.  —  BOOK   IV. 


Probositlon  XV.     Problem. 

385.  To  construct  a  polygon  similar  to  two  given 
similar  polygons  and  equivalent  to  their  difference. 


Al  B'  A  B  A"  B"         P  O 

Let  R  and  R'  be  two  similar  polygons,  and  AB  and 
A'B'  two  homologous  sides. 

To  construct  a  similar  polygon  equivalent  to  H'  —  H. 

Construction.  Construct  the  rt.  Z  F, 

and  take  FO  =  AB. 

From  0  as  a  centre,  with  a  radius  equal  to  A'B', 

describe  an  arc  cutting  P-Z  at  JI,  and  join  OIL 

Take  A".B"  =  FH,  and  on  A"F",  homologous  to  AF, 

construct  F"  similar  to  F. 

Then  F^'  is  the  polygon  required. 

Proof.   FB"  =  0If'-6F',  .'.  A^^W  =  A^  -  IS. 
F'       AJB''' 


Now 


B"     A'^W' 


and  ^^-A£=.  §376 

{similar  polygons  are  to  each  other  as  the  squares  of  their  homologous  sides). 

By  subtraction, 

R'-F^AJF^-AF'__. 
"  F"  A^r^'' 


PEOBLEMS   OF   CONSTRUCTION. 


197 


Proposition  XVI.     Problem. 

386.   To  construct  a  triangle  equivalent  to  a  given 
polygon. 


I       A 


■-STC 


Let  ABCDHE  be  the  given  polygon. 

To  construct  a  triangle  equivalent  to  the  given  polygon. 
Construction,  From  D  draw  DE, 

and  from  ^draw  IIFW  to  DE. 
Produce  AE  to  meet  IFF  at  E,  and  draw  DE. 
Again,  draw  CE,  and  draw  EE  II  to  CE  to  meet  AE  pro- 
duced at  K,  and  draw  CK. 

In  like  manner  continue  to  reduce  the  number  of  sides  of 
the  polygon  until  we  obtain  the  A  CIE. 

Proof.    The  polygon  ABCBE  has  one  side  less  than  the 
polygon  ABCDHE,  but  the  two  are  equivalent. 
For  the  part  ^-SCi)^  is  common, 

and  the  A  DEE^  A  DEH,  §  3G9 

(Jor  the  base  DE  is  common,  and  their  vertices  F  and  JI  are  in  the  line 
FH II  to  the  base). 

The  polygon  ABCK  has  one  side  less  than  the  polygon 
ABODE,  but  the  two  are  equivalent. 

For  the  part  ABCEx?,  common, 
and  the  A  CEK^  A  CED,  §  369 

{^Jor  the  base  CF  is  common,  and  their  vertices  K  and  D  are  in  the  line 
KD  II  to  the  base). 

In  like  manner  the  A  CIK^  ABCK. 

Q.  E.  F. 


198 


PLANE   GEOMETEY.  —  BOOK   IV. 


Proposition  XVII.     Problem. 

387.   To  construct  a  square  which  shall  have  a  given 
ratio  to  a  given  square. 


A"^. 


m- 
11- 


/' 

^<^ 

"-^ 

y' 

X 

/ 

\  \ 

/ 

a 

h    \-( 

IB 

Hi 

~-^^   / 

/ 

E-^«-. 

/ 

n 

•-,y 

Let  R  be  the  given  square,  and  ~  the  given  ratio. 

m 

To  construct  a  square  which  shall  be  to  JR  as  n  is  to  7)i. 

Construction.  Take  AB  equal  to  a  side  of  H,  and  draw  Ai/, 
making  any  acute  angle  with  AB. 

On  At/  take  AU=m,  EF=  n,  and  join  EB. 

Draw  FC  II  to  EB  to  meet  AB  produced  at  C. 

On  J.  (7  as  a  diameter  describe  a  semicircle. 

At  B  erect  the  JL  BD,  meeting  the  semicircumference  at  D. 

Then  BD  is  a  side  of  the  square  required. 

Proof.    Denote  AB  by  a,  BC  by  h,  and  BD  by  x. 

Now  a:x-='x'.h\  that  is,  x^  =  ah.  "§  337 

^^ence,  a^  will  have  the  same  ratio  to  x^  and  to  ao^ 


Therefore  a^  \  a^  —  a^  :  ah  =^  a  :  h. 
But  a  '.  h  ■=  m  :  n, 


§309 


(a  straight  line  drawn  through  two  sides  of  a  A,  parallel  to  the  third  side, 
divides  those  sides  proportionally). 

Therefore  a^ :  x^  =  m  \  n. 


By  inversion,  x^ :  a^ 


n :  m. 


Hence  the  square  on  BD  will  have  the  same  ratio  to  B  as 
n  has  to  m.  q.  e.  f. 


PROBLEMS  OF  CONSTRUCTION.  199 


Proposition  XVIII.     Problem. 

388.   To  construct  a  polygon  similar  to  a  given  poly- 
gon and  having  a  given  ratio  to  it. 


/ 


\ 
\ 
\ 
\ 


V / 

A'  D' 

Let  R  be  the  given  polygon  and  --  the  given  ratio. 

m 

To  construct  a  polygon  shnilar  to  JR,  which  shall  be  to  It  as 
nis  to  m. 

Construction.  Find  a  line  A'B\  such  that  the  square  con- 
structed upon  it  shall  be  to  the  square  constructed  upon  AB 
as  71  is  to  w.  §  387 

Upon  A^Ii'  as  a  side  homologous  to  AB,  construct  the  poly- 
gon S  similar  to  B. 

Then  S  is  the  polygon  required. 

Proof.  S:B  =  A^'':AB',  §376 

{similar  polygons  arc  to  each  other  as  the  squares  of  tlieir  homologous  sides). 

But  A^''':AB''  =  n:m.  Cons. 

Therefore  S:  B  =  n:'m. 

aE.F. 


Ex.  301.  Find  the  area  of  a  right  triangle  if  tlie  length  of  the  hypote- 
nuse is  17  feet,  and  the  length  of  one  leg  is  8  feet. 

Ex.  305.  Compare  the  altitudes  of  two  equivalent  triangles,  if  the 
base  of  one  is  three  times  that  of  the  other. 

Ex.  306.  The  bases  of  a  trapezoid  are  8  feet  and  10  feet,  and  the  alti- 
tude is  6  feet.  Find  the  base  of  an  equivalent  rectangle  having  an  equal 
altitude.  ,J^''  "^ 


200 


PLANE   GEOMETRY. 


BOOK    IV. 


Proposition  XIX.     Problem. 

389.  To  construct  a  square  equivalent  to  a  given 
parallelogram. 


B 


R 


m. 


N  O     ^ 

Let  ABCD  be  a  parallelogram,  b  its  base,  and  a  its 
altitude. 

To  construct  a  square  equivalent  to  the  O  ABCD. 

Oonstmction.   Upon  the  line  MX  take  MJ^=  a,  and  NO  =  b. 

Upon  MO  as  a  diameter,  describe  a  semicircle. 

At  iV  erect  NF  ±  to  MO,  to  meet  the  circumference  at  P. 

Then  the  square  R,  constructed  upon  a  line  equal  to  NP, 
is  equivalent  to  the  O  ABCD. 

Proof.  MN  :NP=NP:  NO,  §  337 

[a  JL  let  fall  from  any  point  of  a  circumference  to  the  diameter  is  a  mean 
proportional  between  the  segments  of  the  diameter). 


That  is, 


.'.NP'=MNxJSrO  = 
P^  a  ABCD. 


axb. 


Q.  E.  F. 


390.  Cor.  1.  A  square  may  he  constructed  equivalent  to  a 
given  triangle,  hy  talcing  for  its  side  a  mean  'pro'portional  be- 
tween the  base  and  one-half  the  altitude  of  the  triangle. 

391.  Cor.  2.  A  square  may  be  constructed  equivalent  to  a 
given  'polygon,  hy  first  reducing  the  -polygon  to  an  equivalent 
triangle^  and  then  constructing  a  square  equivalent  to  the 
triangle. 


a 


PROBLEMS   OF   CONSTRUCTION. 


201 


X  Proposition  XX.     Problem. 

392.  To  construct  a  parallelogram  equivalent  to  a 
^iven  square,  and  having  the  sum  of  its  base  and 
altitude  equal  to  a  given  line. 


R 


W- 


Let  R  be  the  given  square,  and  let  the  sum  of  the 
base  and  altitude  of  the  required  parallelogram  be 
equal  to  the  given  line  MN. 

To  construct  a  O  equivalent  to  H,  with  the  sum  of  its  base 
and  altitude  equal  to  MIT.  ,| 

Oonstraction.  Upon  MN  as  a  diameter,  describe  a  semicircle. 

At  M  erect  a  J.  MP,  equal  to  a  side  of  the  given  square  It. 

Draw  PQ  II  to  MN,  cutting  the  circumference  at  8. 

"Dt^vj  8C 1.  to  MN 

Any  O  having  CM  for  its  altitude  and  CNiov  its  base  is 
equivalent  to  M. 

Proof.  SC=PM.  §§100,180 

.\'8C'  =  PM"=B. 

But  MC:  80=  80:  ON,  §  337 

(a  ±  let  fall  from  any  point  in  the  circumference  to  the  diameter  is  a  mean 
proportional  between  the  segments  of  the  diameter). 

Then  So'^MOxON  a  e.  f. 

Note.  This  problem  may  be  stated :  To  construct  two  straight  lines 
the  sum  and  product  of  which  are  known. 


202 


PLANE   GEOMETRY.  —  BOOK    IV. 


Proposition  XXI.     Problem. 

393.  To  construct  a  parallelograin  equivalent  to  a 
given  square,  and  having  tlie  difference  of  its  base 
and  altitude  equal  to  a  given  line, 

s 


/^ 



/ 

/ 

/ 
/ 

/ 
/ 

n' 

/ 

/ 
/ 
/ 

Let  R  he  the  g:iven  square,  and  let  the  difference  oi 
the  base  and  altitude  of  the  required  parallelogram 
be  equal  to  the  given  line  MN. 

To  construct  a  O  equivalent  to  H,  with  the  difference  of  the 
base  and  altitude  equal  to  MN. 

Oonstmction.  Upon  the  given  line  MN2.^  a  diameter,  describe 
a  circle. 

From  M  draw  MB,  tangent  to  the  O,  and  equal  to  a  side 
of  the  given  square  II . 

Through  the  centre  of  the  O  draw  BB  intersecting  the  cir- 
cumference at  (7  and  B. 

Then  r.ny  O,  as  B) ,  having  BB  for  its  base  and  BC  for  its 
altitude,  is  equivalent  to  R. 

Proof.  BB-BM^BM:  BO,  §  348 

(if  from  a  point  without  aOa  secant  and  a  tangent  are  drawn,  the  tangent  is 
a  mean  proportional  between  the  whole  secant  and  the  part  without  the  O). 

Then  BM'-BBxBC, 

and  the  difference  between  BB  and  BC  is  the  diameter  of  the 
O,  that  is,  J[fiV.  Q.E.F. 

Note.  This  problem  may  be  stated:  To  construct  two  straight  lines 
the  difference  and  product  of  which  are  known. 


PROBLEMS   OF   CONSTRUCTION.  203 


Proposition  XXII.     Problem. 

394.  To  construct  a  polygon  similar  to  a  given  poly- 
gon P,  and  equivalent  to  a  given  polygon  Q, 


\ 

\ 

\ 

\ 
\ 

1 

/ 

/ 

A! 

B' 

7j 

w,. 

A 
Let  P  and  Q  be  two  polygons,  and  AB  a  side  of  P. 

To  construct  a  polygon  similar  to  P  and  equivalent  to  Q. 

Oonstrnction.   Find  squares  equivalent  to  P  and  Q,       §  391 

and  let  m  and  n  respectively  denote  their  sides. 

Find  A'B*,  a  fourth  proportional  to  m,  n,  and  AB.      §  351 

Upon  A'£*,  homologous  to  AB,  construct  P*  similar  to  P. 

Then  P*  is  the  polygon  required. 

Proof.  m:n  —  AB :  A^B*,  Cons 

.-.  nv'-.n'^  AB" :  JJW, 

But  Pom^,  and  Qon^.  Cons, 

,'.  P:  Q^m''  '.n^  =  A^  :  A^'\ 

But  P:P==A^xAJB^,  §376 

{mnilar  "polygons  are  to  each  other  as  the  eqv-ares  of  their  homologous  si-des). 

/.P'.Q^P.r.  Ax.  1 

.'.  P'  is  equivalent  to  Q,  and  is  similar  to  P  by  construction. 


acp. 


^ 


1  HA 


:^ 


204 


PLANE   GEOMETRY.  —  BOOK    IV. 


/ 


Problems  of  Computation. 


Ex.  307.    To  find  the  area  of  an  equilateral  triangle  in  terms  of  its 


side. 


Denote  the  side  by  a,  the  altitude  by  A,  and  the  area  by  S. 
4        4  * 


Then 


Ji'  =  a' 


2 


But 


axh 


2         2 


aVS      a^VS 


Ex.  308.   To  find  the  area  of  a  triangle  in  terms  of  its  sides 
By  Ex.  219 


A  =  7  Vs  (s  —  a)  (s  —  6)  (s  —  c). 

0 


Hence, 


S=^X^  Vs{s-a){s-b){s-c) 
=  Vs  (s  —  a){s  —  b)  (s  —  c). 


Ex.  309.    To  find  the  area  of  a  triangle  in  terms  of  the  radius  of  the 
circumscribing  circle. 

If  B  denote  the  radius  of  the  circumscribing  circle,  and  h  the  altitude 
of  the  triangle,  we  have,  by  Ex.  222, 

bxc  =  2Bxh. 

Multiply  by  a,  and  we  have 

axbxc  =  2Exaxh. 
But  axh  =  2S. 

.•.axbxc  =  4:IiX  S 
abc 


.■.s= 


4:B 


Note.  The  radius  of  the  circumscribing  circle  is  equal  to 


4/Sf 


EXERCISES.         /         i  205 


Theorems. 


^310.  In  a  right  triangle  the  product  of  the  legs  is  equal  to  the  product 
of  the  hypotenuse  and  the  perpendicular  drawn  to  the  hypotenuse  from 
the  vertex  of  the  right  angle. 

\311.  If  ABC  is  a  right  triangle,  C  the  vertex  of  the  right  angle, 
BD  a  line  cutting  AG  in  D,  then  BD"  +  IC'  =  AB'  +  DO". 

"^12.  Upon  the  sides  of  a  right  triangle  as  homologous  sides  three 
similar  polygons  are  constructed.  Prove  that  the  polygon  upon  the 
hypotenuse  is  equivalent  to  the  sum  of  the  polygons  upon  the  legs. 

313.  Two  isosceles  triangles  are  equivalent  if  their  legs  are  equal  each 
to  each,  and  the  altitude  of  one  is  equal  to  half  the  base  of  the  other. 

314.  The  area  of  a  circumscribed  polygon  is  equal  to  half  the  product 
of  its  perimeter  by  the  radius  of  the  inscribed  circle. 

315.  Two  parallelograms  are  equal  if  two  adjacent  sides  of  the  one 
are  equal  respectively  to  two  adjacent  sides  of  the  other,  and  the  included 
angles  are  supplementary. 

316.  Every  straight  line  drawn  through  the  centre  of  a  parallelogram 
divides  it  into  two  equal  parts. 

317.  If  the  middle  points  of  two  adjacent  sides  of  a  parallelogram  are 
joined,  a  triangle  is  formed  which  is  equivalent  to  one-eighth  of  the 
entire  parallelogram. 

318.  If  any  point  within  a  parallelogram  is  joined  to  the  four  vertices, 
the  sum  of  either  pair  of  triangles  having  parallel  bases  is  equivalent  to 
one-half  the  parallelogram. 

319.  The  line  which  joins  the  middle  points  of  the  bases  of  a  trape- 
zoid divides  the  trapezoid  into  two  equivalent  parts. 

320.  The  area  of  a  trapezoid  is  equal  to  the  product  of  one  of  the  legs 
and  the  distance  from  this  leg  to  the  middle  point  of  the  other  leg. 

321.  The  lines  joining  the  middle  point  of  the  diagonal  of  a  quadri- 
lateral to  the  opposite  vertices  divide  the  quadrilateral  into  two  equiva- 
lent parts. 

322.  The  figure  whose  vertices  are  the  middle  points  of  the  sides  of 
any  quadrilateral  is  equivalent  to  one-half  of  the  quadrilateral. 

323.  ABC\b  a  triangle,  M  the  middle  point  of  AB,  P  any  point  in 
AB  between  A  and  M.  If  MD  is  drawn  parallel  to  PC,  and  meeting 
BC  at  D,  the  triangle  BPD  is  equivalent  to  one-half  the  triangle  ABC 


206  PLANE   GEOMETRY.  —  BOOK    IV. 


Numerical  Exercises. 

324.  Find  the  area  of  a  rhombus,  if  the  sum  of  its  diagonals  is  12  feet, 
and  their  ratio  is  3  :  5. 

325.  Find  the  area  of  an  isosceles  right  triangle  if  the  hypotenuse 
is  20  feet. 

326.  In  a  right  triangle,  the  hypotenuse  is  13  feet,  one  leg  is  5  feet. 
Find  the  area. 

327.  Find  the  area  of  an  isosceles  triangle  if  the  base  =  b,  and  leg  =  c. 

328.  Find  the  area  of  an  equilateral  triangle  if  one  side  =  8. 

329.  Find  the  area  of  an  equilateral  triangle  if  the  altitude  =  h. 

330.  A  house  is  40  feet  long,  30  feet  wide,  25  feet  high  to  the  roof, 
and  35  feet  high  to  the  ridge-pole.  Find  the  number  of  square  feet  in 
its  entire  exterior  surface. 

331.  The  sides  of  a  right  triangle  are  as  3  :  4  ;  5.  The  altitude  upon 
the  hypotenuse  is  12  feet.     Find  the  area. 

332.  Find  the  area  of  a  right  triangle  if  one  leg  =  a,  and  the  altitude 
upon  the  hypotenuse  =  h. 

333.  Find  the  area  of  a  triangle  if  the  lengths  of  the  sides  are  104 
feet,  111  feet,  and  175  feet. 

334.  The  area  of  a  trapezoid  is  700  square  feet.  The  bases  are  30  feet 
and  40  feet  respectively.     Find  the  distance  between  the  bases. 

335.  ABCD  is  a  trapezium;  AB^  87  het,  BO  =  119  ieet,  CD  =  41 
feet,  DA  =  169  feet,  A0=  200  feet.     Find  the  area. 

336.  What  is  the  area  of  a  quadrilateral  pircumscribed  about  a  circle 
whose  radius  is  25  feet,  if  the  perimeter  of  the  quadrilateral  is  400  feet  ? 
What  is  the  area  of  a  hexagon  having  an  equal  perimeter  and  circum- 
scribed about  the  same  circle  ? 

337.  The  base  of  a  triangle  is  15  feet,  and  its  altitude  is  8  feet.  Find 
the  perimeter  of  an  equivalent  rhombus  if  the  altitude  is  6  feet. 

338.  Upon  the  diagonal  of  a  rectangle  24  feet  by  10  feet  a  triangle 
equivalent  to  the  rectangle  is  constructed.     What  is  its  altitude  ? 

339.  Find  the  side  of  a  square  equivalent  to  a  trapezoid  whose  bases 
are  56  feet  and  44  feet,  and  each  leg  is  10  feet. 

340.  Through  a  point  Pin  the  side  AB  of  a  triangle  ABC,  a  line  is 
drawn  parallel  to  BC,  and  so  as  to  divide  the  triangle  into  two  equiva- 
lent parts.     Find  the  value  of  AF  in  terms  of  AB. 


EXERCISES.  207 

341.  What  part  of  a  parallelogram  is  the  triangle  cut  off  by  a  line 
drawn  from  one  vertex  to  the  middle  point  of  one  of  the  opposite  sides  ? 

342.  In  two  similar  polygons,  two  homologous  sides  are  15  feet  and 
25  feet.  The  area  of  the  first  polygon  is  450  square  feet.  Find  the  area 
of  the  other  polygon. 

343.  The  base  of  a  triangle  is  32  feet,  its  altitude  20  feet.  What  is 
the  area  of  the  triangle  cut  off  by  drawing  a  line  parallel  to  the  base 
and  at  a  distance  of  15  feet  from  the  base  ? 

344.  The  sides  of  two  equilateral  triangles  are  3  feet  and  4  feet.  Find 
the  side  of  an  equilateral  triangle  equivalent  to  their  sum. 

345.  If  the  side  of  one  equilateral  triangle  is  equal  to  the  altitude  ot 
another,  what  is  the  ratio  of  their  areas  ? 

346.  The  sides  of  a  triangle  are  10  feet,  17  feet,  and  21  feet.  Find 
the  areas  of  the  parts  into  which  the  triangle  is  divided  by  bisecting  the 
angle  formed  by  the  first  two  sides. 

347.  In  a  trapezoid,  one  base  is  10  feet,  the  altitude  is  4  feet,  the  area 
is  32  square  feet  Find  the  length  of  a  line  drawn  between  the  legs 
parallel  to  the  base  and  distant  1  foot  from  it. 

348.  If  the  altitude  A  of  a  triangle  is  increased  by  a  length  m,  how 
much  must  be  taken  from  the  base  a  in  order  that  the  area  may  remain 
the  same  ? 

349.  Find  the  area  of  a  right  triangle,  having  given  the  segments  p, 
q,  into  which  the  hypotenuse  is  divided  by  a  perpendicular  drawn  to  the 
hypotenuse  from  the  vertex  of  the  right  angle. 

Problems. 

350.  To  construct  a  triangle  equivalent  to  a  given  triangle,  and 
having  one  side  equal  to  a  given  length  I. 

351.  To  transform  a  triangle  into  an  equivalent  right  triangle. 

352.  To  transform  a  triangle  into  an  equivalent  isosceles  triangle. 

353.  To  transform  a  triangle  ABO  into  an  equivalent  triangle,  hav- 
ing one  side  equal  to  a  given  length  I,  and  one  angWequal  to  angle  BAC. 

Hints.  Upon  AB  (produced  if  necessary),  take  AD  =  I,  draw  BE  II  to 
CD,  and  meeting  J. C  (produced  if  necessary)  at  -E";  A  BED=o=ABEC. 

354.  To  transform  a  given  triangle  into  an  equivalent  right  triangle, 
having  one  leg  equal  to  a  given  length. 


208  PLANE   GEOMETRY. BOOK   IV. 

355.  To  transform  a  given  triangle  into  an  equivalent  right  triangle, 
having  the  hypotenuse  equal  to  a  given  length. 

356.  To  transform  a  given  triangle  into  an  equivalent  isosceles  tri- 
angle, having  the  base  equal  to  a  given  length. 

To  construct  a  triangle  equivalent  to : 
357.'  The  sum  of  two  given  triangles. 

358.  The  difference  of  two  given  triangles. 

359.  To  transform  a  given  triangle  into  an  equivalent  equilateral 
triangle. 

To  transform  a  parallelogram  into  : 

360.  A  parallelogram  having  one  side  equal  to  a  given  length. 

361.  A  parallelogram  having  one  angle  equal  to  a  given  angle. 

362.  A  rectangle  having  a  given  altitude. 
To  transform  a  square  into  : 

363.  An  equilateral  triangle. 

364.  A  right  triangle  having  one  leg  equal  to  a  given  length. 

365.  A  rectangle  having  one  side  equal  to  a  given  length. 

To  construct  a  square  equivalent  to : 

366.  Five-eighths  of  a  given  square. 

367.  Three-fifths  of  a  given  pentagon. 

368.  To  draw  a  line  through  the  vertex  of  a  given  triangle  so  as  to 
divide  the  triangle  into  two  triangles  which  shall  be  to  each  other  as  2 :  3. 

369.  To  divide  a  given  triangle  into  two  equivalent  parts  by  drawing 
a  line  through  a  given  point  P  in  one  of  the  sides. 

370.  To  find  a  point  within  a  triangle,  such  that  the  lines  joining  this 
point  to  the  vertices  shall  divide  the  triangle  into  three  equivalent  parts. 

371.  To  divide  a  given  triangle  into  two  equivalent  parts  by  drawing 
a  line  parallel  to  one  of  the  sides. 

372.  To  divide  a  given  triangle  into  two  equivalent  parts  by  drawing 
a  line  perpendicular  to  one  of  the  sides. 

373.  To  divide  a  given  parallelogram  into  two  equivalent  parts  by 
drawing  a  line  through  a  given  point  in  one  of  the  sides. 

374.  To  divide  a  given  trapezoid  into  two  equivalent  parts  by  draw- 
ing a  line  parallel  to  the  bases. 

375.  To  divide  a  given  trapezoid  into  two  equivalent  parts  by  draw- 
ing a  line  through  a  given  point' in  one  of  the  bases. 


J<jyv 


LO^^^/^ 


BOOK   V. 
REGULAR   POLYGONS  AND   CIRCLES. 

395.  A  regular  polygon  is  a  polygon  which  is  equilateral 
and  equiangular ;  as,  for  example,  the  equilateral  triangle,  and 
the  square. 

Proposition  I.    Theorem. 

396.  An  equilateral  polygon  inscribed  in  a  circle  is 
a  regular  jx)lygon. 


Let  ADC,  etc.,  be  an  equilateral  polygon  inscribed  in 
a  circle. 

To  prm>e  the  polygon  ABC,  etc.,  regular. 

Proof.        The  arcs  AB,  BC,  CD,  etc.,  are  equal,  §  230 

(in  the  same  O,  equal  chords  subtend  equal  arcs). 

Hence  arcs  ABC,  BCD,  etc.,  are  equal.  Ax.  6 

and  the  A  A,  B,  C,  etc.,  are  equal, 

{being  inscribed  in  equal  segments). 

Therefore  the  polygon  ABC,  etc.,  is  a  regular  polygon,  being 
equilateral  and  equiangular.  q.e.  d. 


210  PLANE    GEOMETRY.  —  BOOK    V. 


Proposition  II.     Theorem. 

397.  A  circle   may  he   circurnscrihed   about,  and  a 
circle  may  he  inscrihed  in,  any  regular  polygon. 


Let  ABODE  be  a  regular  polygon. 

I.    To   prove    that   a   circle   may   be   circumscribed  about 
ABODE. 

Proof.    Let  0  be  tlie  centre  of  the  circle  passing  through 
A,B,C, 

Join  OA,  OB,  00,  and  OD. 

Since  the  polygon  is  equiangular,  and  tlie  A  OBOis  isosceles, 

Z.ABO=^Z.BOD 

and  Z.OBO.^4-  OOB 

By  subtraction,        Z.  OB  A  =  Z  OCD 

Hence  in  the  A  OB  A  and  OCD 

the  Z.  OB  A  =  /.  OCD, 

the  radius  OB  =  the  radius  OC, 

and  AB--=CD.  §395 

.•.AOAB  =  AOCD,  '     §150 

{having  two  sides  and  the  included  /.  of  the  one  equal  to  two  sides  and  the 
included  Z  of  the  other). 

.-.  OA  =  OD. 
Therefore  the  circle  passing  through  A,  B,  and   C,  also 
passes  through  D. 


REGULAR   POLYGONS   AND   CIRCLES.  211 

In  like  manner  it  may  be  proved  that  the  circle  passing 
through  B,  C,  and  D,  also  passes  through  E\  and  so  on 
through  all  the  vertices  in  succession. 

Therefore  a  circle  described  from  0  as  a  centre,  and  with  a 
radius  OA,  will  be  circumscribed  about  the  polygon. 

II.    To  prove  that  a  circle  may  he  inscribed  in  ABODE. 

Proof,  Since  the  sides  of  the  regular  polygon  are  equal 
chords  of  the  circumscribed  circle,  they  are  equally  distant 
from  the  centre.  §  236 

Therefore  a  circle  described  from  0  as  a  centre,  and  with 
the  distance  from  0  to  a  side  of  the  polygon  as  a  radius,  will 
be  inscribed  in  the  polygon.  aE.o. 

398.  The  radius  of  the  circumscribed  circle,  OA,  is  called 
the  radius  of  the  polygon. 

399.  The  radius  of  the  inscribed  circle,  OF,  is  called  the 
apothem  of  the  polygon. 

400.  The  common  centre  0  of  the  circumscribed  and  in- 
scribed circles  is  called  the  centre  of  the  polygon. 

401.  The  angle  between  radii  drawn  to  the  extremities  of 
any  side,  as  angle  AOB,  is  called  the  angle  at  the  centre  of  the 
polygon. 

By  joining  the  centre  to  the  vertices  of  a  regular  polygon, 
the  polygon  can  be  decomposed  into  as  many  equal  isosceles 
triangles  as  it  has  sides.     Therefore, 

402.  Cor.  1.  The  angle  at  the  centre  of  a  regular  polygon  is 
equal  to  four  right  angles  divided  by  the  number  of  sides  of 
the  polygon. 

403.  Cor.  2.  The  radius  drawn  to  any  vei-tex  of  a  regular 
polygon  bisects  the  angle  at  the  vertex. 

404.  Cor.  3.  jThe  interior  angle  of  a  regular  polygon  is  the 
supplement  of  the  angle  at  the  centre^ 

For  the  Z  AB0=2ZAB0={Z  ABO-\-ZBAO.  Hence 
the  A  ABO  is,  the  supplement  of  tne  Z  AOB. 


212  PLANE   GEOMETRY. BOOK    V. 

Proposition  III.     Theorem. 

/  405.  If  the  circumference  of  a  circle  is  divided  into 
any  number  of  equal  parts,  the  chords  joining  the 
successive  points  of  division  form  a  regular  inscribed 
polygon,  and  the  tangents  drawn  at  the  points  of 
division  form  a  regular  circumscribed  polygon. 
T         D         H 


F 
Let  the  circumference  be  divided  into  equal  arcs, 
AB,  BC,  CD,  etc.,  be  chords,  FBG,  GGH,  etc.,  be  tangents. 

I.  To  prove  that  ABCDE  is  a  regular  polygon. 

Proof.       The  sides  AB,  BC,  CD,  etc.,  are  equal,  §  230 

{in  the  same  O  equal  arcs  are  subtended  by  equal  chords). 

Therefore  the  polygon  is  regular,  §  396 

{an  equilateral  polygon  inscribed  in  a  O  is  regular). 

II.  To  prove  that  the  polygon  FGHIK  is  a  regular  polygon. 
Proof.   In  the  A  AFB,  BOC,  CHD,  etc. 

AB  =  BC=  CD,  etc.  §  395 

Also,  Z  BAF=  Z.  ABF=  Z  CBG  =  Z  BCG,  etc.,    §  269 
{being  measured  by  halves  of  equal  arcs). 

Therefore  the  triangles  are  all  equal  isosceles  triangles. 

Hence  ZF=ZQ  =  AH,  etc. 

Aiso,  FB^BG^GC  =  CH,  etc. 

Therefore  i^6^  =  6«.Zr,  etc. 

/.  FGHIK  IB  a  regular  polygon.  §  395 

aE.  D. 
406.   Cor.  1.   Tangents  to  a  circumference  at  the  vertices  of  a 
regular  inserihed  polygon  form  a  regular  circumscribed  poly- 
gon  of  the  same  number  of  sides. 


REGULAR   POLYGONS   AND   CIRCLES. 


213 


407.  Cor.  2.  If  a  regular  polygon  is  inscribed  in  a  circle, 
the  tangents  drawn  at  the  middle  points 
of  the  arcs  subtended  by  the  sides  of  the 
polygon  form  a  circumscribed  regular 
polygon,  whose  sides  are  parallel  to  the 
sides  of  the  inscribed  polygon  and  whose 
vertices  lie  on  the  radii  {prolonged)  of 
the  inscribed  polygon.  For  any  two  cor-  A  M  n' 
responding  sides,  as  AB  and  A'B\  perpendicular  to  OM, 
are  parallel,  and  the  tangents  MB^  and  NB\  intersecting  at  a 
point  equidistant  from  OJf  and  OiV(§  246),  intersect  upon  the 
bisector  of  the  Z  MOJSr(§  163) ;  that  is,  upon  the  radius  OB. 

408.  Cor.  3.  ^  the  vertices  of  a  regular  inscribed  polygon 
are  joined  to  the  middle  points  of  the  arcs  sub- 
tended by  the  sides  of  the  polygon,  the-  joining 
lines  form   a  regular   inscribed  polygon    of 
double  the  number  of  sides. 

409.  Cor.  4.  If  tangents  are  drawn  at  the 
middle  points  of  the  arcs  between  adjacent 
points  of  contact  of  the  sides  of  a  regular  cir- 
cumscribed polygon,  a  regular  circumscribed 
polygon  of  double  the  number  of 
formed. 

410.  Scholium.  The  perimeter  of  an  inscribed  polygon  is 
less  than  the  perimeter  of  the  inscribed  polygon  of  double  the 
number  of  sides;  for  each  pair  of  sides  of  the  second  polygon 
is  greater  than  the  side  of  the  first  polygon  which  they  replace 
(§137). 

The  perimeter  of  a  circumscribed  polygon  is  greater  than 
the  perimeter  of  the  circumscribed  polygon  of  double  the  num- 
ber of  sides ;  for  every  alternate  side  FO,  SI,  etc.,  of  the  poly- 
gon FGIII,  etc.,  replaces  portions  of  two  sides  of  the  circum- 
scribed polygon  ABCD,  and  forms  with  them  a  triangle,  and 
one  side  of  a  triangle  is  less  than  the  sum  of  the  other  two  sides. 


214  PLANE   GEOMETRY.         BOOK   V. 


Proposition  IV.     Theorem. 


li 


411.  Two  regular  polygons  of  the  same  nwmber  of 
sides  are  similar. 


A!  n 

Let  Q  and  Q'  be  two  regular  polygons,  each  having 
n  sides. 

To  prove  Q  and  Q'  similar  polygons. 

Proof.  The  sum  of  the  interior  A  of  each  polygon  is  equal  to 

(n~2)2  rt.  A,  §  205 

{the  sum  of  the  interior  A  of  a  polygon  is  equal  to  2  rt.  A  taken  as  many 
times  less  2  as  the  polygon  has  ddes). 

Each  angle  of  either  polygon  =  ^ ^ — '- — >     §  206 

{for  the  A  of  a  regular  polygon  are  all  equal,  and  hence  each  Z  is  equal 
to  the  sum  of  the  A  divided  by  their  number). 

Hence  the  two  polygons  Q  and  Q'  are  mutually  equiangular. 

Since  AB  =  BC,  etc.,  and  A'B'  =  B'G\  etc.,  §  395 

AB\A'B'=^BC\B'Q\^tQ, 

Hence  the  two  polygons  have  their  homologous  sides 
proportional. 

Therefore  the  two  polygons  are  similar.  §  319 

a  E.D. 

412.  Cob.  The  areas  of  two  regular  polygons  of  the  same 
number  of  sides  are  to  each  other  as  the  squares  of  any  two 
homologous  sides.  §  376 


REGULAR   POLYGONS   AND   CIRCLES. 


215 


Proposition  V.     Theorem. 

-  -413.  The  perimeters  of  two  regular  polygons  of  the 
same  number  of  sides  are  to  each  otJier  as  the  radii 
of  tlwir  eircumscrihed  circles,  and  also  as  the  radii 
of  their  inscribed  circles. 


A     M    n  ^i'      M'     B' 

Let  P  and  P'  denote  the  perimeters,  0  and  O  the 
centres,  of  the  two  regular  polygons. 

From  0,  0'  draw  OA,  O'A',  OB,  O'B',  and  Js  OM,  O'M'. 

To  prove      F :  P  =  OA  :  O'A'  =  OM:  O'M'. 

Proof.    Since  the  polygons  are  similar, 

P:P  =  AB:A'B'. 

In  the  isosceles  A  OAB  and  O'A'B' 

the  Z  0  =  the  Z  0', 

and  OA  -'OB  =  O'A' :  O'B'. 

.-.  the  A  OAB  and  O'A'B'  are  similar. 

.-.  AB  :  A'B'  =  OA  :  O'A', 

Also  AB  :  A'B'  =  OM:  O'M', 

{the  homologous  altitudes  of  similar  ^  have  the  same  ratio  as  their  bases). 

.-.  F:F'=OA:  O'A'^OM:  0M\ 

aE,  D. 

414.  Cor.  The  areas  of  two  regular  polygons  of  the  same 
number  of  sides  are  to  each  other  as  the  squares  of  the  radii 
of  their  drcumseribed  circles,  and  also  as  the  squares  of  the 
radii  of  their  inscribed  circles.  §  376 


§411 
§333 

§402 

§326 
§319 

§328 


216 


PLANE   GEOMETKY. 


BOOK    V. 


Proposition  VI.     Theorem. 

415.  The  difference  hetween  the  lengths  of  the  perim- 
eters of  a  regular  inscribed  polygon  and  of  a  similar 
circumscrihed  polygon  is  indefinitely  diminished  as 
the  number  of  the  sides  of  the  polygons  is  indefinitely 
increased. 


Let  P  and  P'  denote  the  lengths  of  the  perimeters, 
AB  and  A'B'  two  corresponding'  sides,  OA  and  OA'  the 
radii,  of  the  polygons. 

To  prove  that  as  the  number  of  the  sides  of  the  polygons  is 
indefinitely  increased,  JP  ~  F  is  indefinitely  diminished. 

Proof.   Since  tlie  polygons  are  similar, 
F:P=OA':OA. 
By  division,    P~  P-.  P=  OA^  -  OA:  OA. 

OA'  -  OA 


§413 


Whence 


F-P=PX 


OA 


Draw  the  radius  OC  to  the  point  of  contact  of  A'B\ 

In  the  A  OA'C,       OA'  ~OC<A'0,  §  137 

(the  difference  between  two  sides  of  a  A  is  less  than  the  third  side). 

Substituting  OA  for  its  equal  00, 

OA'-~OA<A'0. 

But  as  the  number  of  sides  of  the  polygon  is  indefinitely 
increased,  the  length  of  each  side  is  indefinitely  diminished  ; 
that  is,  A'P',  and  consequently  A'O,  is  indefinitely  diminished. 


KEGULAE,  POLYGONS  AND  CIRCLES.        217 

Therefore  OA'—OA,  which  is  less  than  A'C,  is  indefinitely 

diminished;  and  the  fraction — ,  the  denominator  of 

OA 

which  is  the  constant  OA,  is  indefinitely  diminished. 
But  P  always  remains  less  than  the  circumference. 
Therefore  jP— Pis  indefinitely  diminished.  q. e. d. 

416.  Cor.  The  diffei-ence  between  the  areas  of 'a  regular 
insci'ibed  polygon  and  of  a  similar  circumsciibed  polygon  is 
indefinitely  diminished  as  the  number  of  the  sides  of  the  poly- 
gons is  indefinitely  increased. 

For,  if  ;iS'  and  S'  denote  the  areas  of  the  polygons, 

S' :  S=(6A''  :  Oa)  =  OA'" :  OC".  §  414 

By  division,  S'-8:S=  OA^  -  OC' :  0C\ 

Whence    S' - 8=  8x^^^^^£^  =  8x^'^ 


OC"  00" 

Since  A^O  can  be  indefinitely  diminished  by  increasing  the 
number  of  the  sides,  &  —  8  can  be  indefinitely  diminished. 

417.  Scholium.  The  perimeter  P'  is  constantly  greater 
than  P,  and  the  area  8^  is  constantly  greater  than  8\  for  the 
radius  OA^  is  constantly  greater  than  OA.  But  P'  constantly 
decreases  and  P  constantly  increases  (§  410),  and  the  area  xS" 
constantly  decreases,  and  the  area  8  constantly  increases,  as 
the  number  of  sides  of  the  polygons  is  increased. 

Since  the  difference  between  P'  and  P  can  be  made  as 
small  as  we  please,  but  cannot  be  made  absolutely  zero,  and 
since  P'  is  decreasing  while  P  is  increasing,  it  is  evident  that 
P'  and  P  tend  towards  a  common  limit.  This  common  limit 
is  the  length  of  the  circumfe^-ence.  §  259 

Also,  since  the  difference  between  the  areas  8^  and  8  can  be 
made  as  small  as  we  please,  but  cannot  be  made  absolutely 
zero,  and  since  8^  is  decreasing,  while  8  is  increasing,  it  is 
evident  that  /S"  and  8  tend  towards  a  common  limit.  This 
common  limit  is  the  area  of  the  circle. 


218 


PLANE   GEOMETRY.  —  BOOK   V. 


Proposition  VII.     Tiieoeem. 

418.  Two  circinjvferences  have  the  same  ratio  ad 
their  radii. 


Let  C  and  C  be  the  circumferences,  R  and  B'  the 
radii,  of  the  two  circles  Q  and  Q'. 

To  prove  Q\Q^=-R\R\ 

Proof,  Inscribe  in  the  (D  two  similar  regular  polygons,  and 
denote  their  perimeters  by  P  and  P\ 

Then  P'.F  =  R'.R\%  413)  ;  that  is,  R^xP^RxP. 

Conceive  the  number  of  the  sides  of  these  similar  regular 
polygons  to  be  indefinitely  increased,  the  polygons  continuing 
to  have  an  equal  number  of  sides. 

Then  R'  X  P  will  continue  equal  to  Rx  P,  and  P  and  P 
will  approach  indefinitely  C  and  C  as  their  respective  limits. 

.-.  R^X  C^Rx  C'(§260);  that  is,  O:  C  =  R  :  R\ 

Q.  E.  O. 

419.  Cor.  The  ratio  of  the  circumference  of  a  circle  to  its 
diameter  is  constant.  For,  in  the  above  proportion,  by  doubling 
both  terms  of  the  ratio  R  :  R\  we  have 

C:C'    =2R'.2R\ 
By  alternation,        0'.2R=C'    :2R\ 
Thi«  constant  ratio  is  denoted  by  tt,  so  that  for  any  circle 
whose  diameter  is  2  i?  and  circumference  (7,  we  have 

— -TT,  or  C=27ri^. 
2R 

420.  Scholium.  The  ratio  tt  is  incommensurable,  and  there- 
fore can  be  expressed  in  figures  only  approximately. 


REGULAR  POLYGONS  AND  CIRCLES.        219 


Proposition  VIII.     Theorem. 

421.   The  area  of  a  regular  polygon  is  equal  to  one- 
half  the  product  of  its  apothem  by  its  periineter. 


Let  P  represent  the  perimeter,  R  the  apothem,  and 
S  the  area  of  the  regular  polygon  ABC,  etc. 

To  prove  S^^BxR 

Proof.  Draw  OA,  OB,  00,  etc. 

The  polygon  is  divided  into  as  many  A  as  it  has  sides. 

The  apothem  is  the  common  altitude  of  these  A, 

and  the  area  of  each  A  is  equal  to  ^  JR  multiplied  by  tlie 
base.  ^  §  368 

Hence  the  are.*  of  all  the  A  is  equal  to  \Ii  multiplied  by 
the  sum  of  all  the  bases. 

But  the  sum  of  the  areas  of  all  the  A  is  equal  to  the  area 
of  the  polygon, 

and  the  sum  of  all  the  bases  of  the  A  is  equal  to  the  perim- 
eter of  the  polygon. 

Therefore /S' =  4  i^X  P. 

aE.  o. 

422.  In  different  circles  similar  arcs,  similar  sectors,  and 
similar  segments  are  such  as  correspond  to  equal  angles  at 
the  centre. 


220  PLANE   GEOMETRY. — BOOK  V. 


Proposition  IX.     Theorem. 

423.  The  area,  of  a  eirele  is  equal  to  one-half  the 
product  of  its  radius  by  its  circumference. 


J 

B         M         C 
Let  li   represent   the  radius,    G  the  circumference, 
and  S  the  area,  of  the  circle. 

To  prove  8-=\BxC. 

Proof.  Circumscribe  any  regular  polygon  about  the  circle, 
and  denote  its  perimeter  by  P. 

Then  the  area  of  this  polygon  =  ^Rx  F,       §  421 

Conceive  the  number  of  sides  of  the  polygon  to  be  indefi- 
nitely increased ;  then  the  perimeter  of  the  polygon  aj)proaches 
the  circumference  of  the  circle  as  its  limit,  and  the  area  of  the 
polygon  approaches  the  circle  as  its  limit. 

But  the  area  of  the  polygon  continues  to  be  equal  to  one- 
half  the  product  of  the  radius  by  the  perimeter,  however  great 
the  number  of  sides  of  the  polygon. 

Therefore  ^-li^X  a  §260 

Q.  E.  D. 

424.  Cor.  1.  The  area  of  a  sector  equals  one-half  the  product 
of  its  radius  by  its  arc.  For  the  sector  is  such  a  part  of  the 
circle  as  its  arc  is  of  the  circumference. 

425.  Cor.  2.  The  area  of  a  circle  equals  tr  times  the  square 
of  its  radius. 

For  the  area  of  the  O -^  \  Ex  C=^  Rx2'7rR  =  irE\ 


REGULAR  POLYGONS  AND  CIRCLES.        221 

426.  Cor.  3.  The  areas  of  two  circles  are  to  each  other  as  the 
squares  of  their  radii.  For,  if  8  and  /S"  denote  the  areas,  and 
M  and  i?'  the  radii, 

8:  8'  ^irE"  :7rR''  =  B" :  R'\ 

427.  Cor.  4.  8imilar  arcs,  being  like  parts  of  their  respective 
circumferences,  are  to  each  other  as  their  radii;  similar  sectoo^s, 
being  like  parts  of  their  respective  circles,  are  to  each  other  as 
the  squares  of  their  radii. 

Proposition  X.    Theorem. 

428.  The  areas  of  two  similar  segments  are  to  each 
other  as  the  squares  of  their  radii. 


P  P' 

Let  AC  and  A'C  be  the  radii  of  the  two  similar  seg- 
ments ABP  and  A'B'P'. 

To  prove         ABP :  A'B'P  =  AC' :  AJW. 

Proof.    The  sectors  ACB  and  A*Q'B*  are  similar,         §  422 
{having  the  A  at  the  centre,  C  and  (7,  equal). 

Ini^iQ  A  ACB  ?ii\d.  A'C'B' 

/.  0=  A  C\  A0=  CB,  and  A'C  =  C'B'. 
Therefore  the  A  ACB  and  A' C'B'  are  similar.    §  326 
Now     sector  A  CB :  sector  A' C'B'  =  AO" :  A^'\       §  427 
and  AACB:AA'C'B'  =  AG^:AJG^i\  §375 

Hence  ^^otor  ACB  -  A  ACB    ^Jg, 

sector  A' C'B'  -  A  A' C'B'     JTq,^  ^ 


That  is,         ABF :  A'B'F^  =  AC^ :  A'C' 


Q.E.D. 


222 


PLANE   GEOMETRY.  —  BOOK    V. 


Problems  of  Construction. 
Proposition  XI.     Problem. 
429.  To  inscribe  a  square  in  a  given  circle. 


Let  0  be  the  centre  of  the  given  circle. 

To  inscribe  a  square  in  the  circle. 

Construction.    Draw  the  two  diameters  AC  and  BD  X  to 
each  other. 

Join  AB,  BC,  CD,  and  DA. 

Then  ABCD  is  the  square  required. 

Proof.  The  A  ABC,  BCD,  etc.,  are  rt.  A,  §  264 

{heing  inscribed  in  a  semicircle), 

and  the  sides  AB,  BC,  etc.,  are  equal,  §  230 

{in  the  same  O  equal  arcs  are  subtended  by  equal  chords). 

Hence  the  figure  ABCD  is  a  square. 


§171 

Q.  E.  F. 


430.  Cor.  By  bisecting  the  arcs  AB,  BC,  etc.,  a  regular 
polygon  of  eight  sides  may  be  inscribed  in  the  circle ;  and,  by 
continuing  the  process,  regular  polygons  of  sixteen,  thirty-two, 
sixty-four,  etc.,  sides  may  be  inscribed. 


Ex.  376.   The  area  of  a  circumscribed  square  is  equal  to  twice  the 
area  of  the  inscribed  square. 

Ex.  377.    If  the  length  of  the  side  of  an  inscribed  square  is  2  inches,    / 
what  is  the  length  of  the  circumscribed  square  ?  / 


PROBLEMS   OF   CONSTRUCTION.  223 

Proposition  XII.     Problem. 
431.  To  inscribe  a  regular  hexagon  in  a  given  circle. 


Let  0  be  the  centre  of  the  given  circle. 

To  inscribe  in  the  given  circle  a  regular  hexagmi. 
Construction.    From  0  draw  any  radius,  as  OC. 

From  Cas  a  centre,  with  a  radius  equal  to  0(7, 
describe  an  arc  intersecting  the  circumference  at  F. 

Draw  Oi^  and  CF. 
Then  CF  is  a  side  of  the  regular  hexagon  required. 
Proof.   The  A  OFQ  is  equilateral  and  equiangular. 
Hence  the  Z  FOC  is  \  of. 2  rt.  A,  or  |  of  4  rt.  A.          §  138 
And  the  arc  i^Cis  \  of  the  circumference  ABCF. 
Therefore  the  chord  FC,  which  subtends  the  arc  FC,  is  a 
side  of  a  regular  hexagon  ; 

and  the  figure  CFD,  etc.,  formed  by  applying  the  radius  six 
times  as  a  chord,  is  a  regular  hexagon.  q.  e.  p. 

432.  Cor.  1.  By  joining  the  alteimaie  voiices  A,  C,  D,  an 
equilate^'al  triangle  is  inscribed  in  the  circle. 

433.  Cor.  2.  By  bisecting  the  arcs  AB,  BC,  etc.,  a  regular 
polygon  of  twelve  sides  may  be  inscribed  in  the  circle ;  and,  by 
continuing  the  process,  regular  polygons  of  twenty-four,  forty- 
eight,  etc.,  sides  may  be  inso'ibed. 


224  PLANE   GEOMETRY.  —  BOOK    V. 

Proposition  XIII.    Problem. 
434.  To  inscribe  a  regular  decagon  in  a  given  circle. 


n 

Let  0  be  the  centre  of  the  given  circle. 

To  inscnhe  a  regular  decagon  in  the  given  circle. 

Oonstruction.  Draw  the  radius  OC, 

and  divide  it  in  extreme  and  mean  ratio,  so  that  00  shall 
be  to  08  as  08  is  to  80.  §  355 

From  (7  as  a  centre,  with  a  radius  equal  to  08, 

describe  an  arc  intersecting  the  circumference  at  B,  and 
draw  BO. 

Then  ^C^is  a  side  of  the  regular  decagon  required. 

Proof.  Draw  B8  and  BO. 

By  construction       00:08=08:80, 

and  B0=  08. 

.:  00:BO=BO:80 

Moreover,  the  Z  OOB  =  Z  80B.  Iden. 

Hence  the  A  OOB  and  B08  are  similar,         §  32G 
{having  an  Z  of  the  one  equal  to  an  Z  of  the  other,  and  the  including  sides 
proportional). 

But  the  A  OOB  is  isosceles, 

(its  sides  OC  and  OB  being  radii  of  the  same  circle). 

.'.  A  B08,  which  is  similar  to  the  A  OOB,  is  isosceles, 

and  0B  =  B8=08 


PROBLEMS   OF   CONSTRUCTION.  225 

/.  the  A  80B  is  isosceles,  and  the  Z  O  =  Z  8B0. 

But  the  ext.  Z  CSB^-Z  0  +  /.8B0  =  2Z  0.    §  145 

Hence        .     Z  8CB  (=  Z  C8B)  =  2  Z  0,  §  154 

and  Z  05(7(=  Z  /S'OS)  =  2  Z  0.  §  154 

.-.  the  sum  of  the  A  of  the  A  OCB  =  5  Z  0  =  2  rt.  A, 

and  Z  0  -:  -^  of  2  rt.  ^,  or  yV  of  4  rt.  ^. 

Therefore  the  arc  BC\&  -^^  of  the  circumference, 

and  the  chord  BO  is  a  side  of  a  regular  inscribed  decagon. 

Hence,  to  inscribe  a  regular  decagon,  divide  the  radius  in 

extreme  and  mean  ratio,  and  apply  the  greater  segment  ten 

times  as  a  chord. 

aE.  F. 

435.  Cor.  1.  Bi/  joining  the  alternate  vertices  of  a  regular 
inscribed  decagon,  a  regular  pentagon  is  inscribed. 

436.  Cor.  2.  Bi/  bisecting  the  arcs  BC,  OF,  etc.,  a  regular 
polygon  of  twenty  sides  may  be  insci^ibed;  and,  by  continuing 
the  process,  regular  polygons  of  forty,  eighty,  etc.,  sides  may  be 
inscribed. 


Let  R  denote  the  radius  of  a  regular  inscribed  polygon,  r  the  apothem, 
a  one  side,  A  an  interior  angle,  and  C  the  angle  at  the  centre ;  show  that 

Ex.  378.   In  a  regular  inscribed  triangle  a  =  R  \/3,  r  =  J  jf2,  ^  =  60°, 
C=  120°. 

Ex.379.   In  an   inscribed  square   a  =  BV2,  r^^EVI,  k  =  90°, 
)o=  90°. 

'  Ex.  380.   In  a  regular  inscribed  hexagon  )i  =  R,r  =  iR  Vs,  ^  =  1 20° 
"^=60°. 

Ex.  381.   In  a  regular  inscribed  decagon 

\a  =  ^(^~^),    r  =  \RVlO  +  2V5,   ^  =  144°,    C=36°. 


226  PLANE   GEOMETRY.  —  BOOK   V. 


Proposition  XIV.     Problem. 

437.  To  inscribe  in  a  given ,  circle  a  regular  pente* 
decagon,  or  polygon  of  fifteen  sides. 


F 
Let  Q  be  the  given  circle. 

To  inscribe  in  Q  a  regular  pentedecagon. 

Construction.  Draw  -E'ZT  equal  to  a  side  of  a  regular  inscribed 
hexagon,  §  431 

and  ^i^ equal  to  a  side  of  a  regular  inscribed  decagon.  §  434 

Join  FH. 

Then  FII  will  be  a  side  of  a  regular  inscribed  pentedecagon. 

Proof.        The  arc  EH  is  ^  of  the  circumference, 

and  the  arc  EF  is  -^  of  the  circumference. 

Hence  the  arc  FH  is  |-  —  y^^,  or  -^,  of  the  circumference, 

and  the  chord  FH  is  a  side  of  a  regular  inscribed  pente- 
decagon. 

By  applying   FH  fifteen  times  as  a  chord,  we  have  the 

polygon  required. 

a  E.  F. 

438.  Cor.  By  bisecting  the  arcs  FH,  HA,  etc.,  a  regular 
polygon  of  thirty  sides  may  be  inscribed;  and,  by  continuing 
the  process,  regular  polygons  of  sixty,  one  hundred  and  twenty^ 
etc.,  sides,  may  be  inscribed. 


PROBLEMS   OF   CONSTRUCTION.  227 


Proposition  XV.     Problem. 

439.  To  inscribe  in  a  given  circle  a  regular  polygon 
similar  to  a  given  regular  polygon. 


Let  ABCD,  etc.,  be  the  given  regular  polygon,  and 
C'D'E'  the  given  circle. 

To  inscribe  in  the  circle  a  regular  polygon  similar  to  ABCD, 
etc. 

Oonstniction.   From  0,  the  centre  of  the  given  polygon, 

draw  OD  and  OC. 

From  0',  the  centre  of  the  given  circle, 

draw  aC^  and  aD', 

making  the  Z  0'=Z  0. 

Draw  CD*. 

Then  CD'  will  be  a  side  of  the  regular  polygon  required. 

Proof.    Each  polygon  will  have  as  many  sides  as  the  A  0 
(=  Z  0')  is  contained  times  in  4  rt.  A. 

Therefori^the  polygon  CD'U',  etc.,  is  similar  to  the  poly- 
gon CDE,  etc.,  §  411 

{tvjo  regular  polygons  of  the  same  number  of  sides  are  similar). 


228 


PLANE   GEOMETRY.  —  BOOK    V. 


Proposition  XVI.     Problem. 

440.  Given  the  radius  and  the  side  of  a  regular 
inscribed  polygon,  to  find  the  side  of  the  regular 
inscribed  polygon  of  double  the  number  of  sides. 

^' 


LetAB  he  a  side  of  the  regular  inscrihed  polygon. 

To  find  the  value  of  AD,  a  side  of  a  regular  inscribed  poly- 
gon of  double  the  number  of  sides. 

From  D  draw  i)-9" through  the  centre  O,  and  draw  OA,  AH. 
BJIis  ±  to  AB  at  its  middle  point  O.  §  123 

Inthert.AO^a  00^=  OT 
That  is, 
But 


A0\ 


§339 


OC=^OA^-^. 
AO=^AB;  hence  Ae^^-i-AW' 

Therefore,  0C=-^6A" -^AB". 

In  the  rt.A  BAIT, 

AB'  =  B2IxBC 

=  2  0A(0A-0C), 
and  AB  =  ^/20A(0A- 00). 


§264 
§334 


If  we  denote  the  radius  by  B,  and  substitute  '\/ B^  —  \A]^ 
for  00,  then         /X 

AB  =  '\l2B(B~-^B''-i^)    jlr"' 
=  A/i^(2  B-  ^'^R^-AS). 


Q.C.F. 


PROBLEMS   OF   COMPUTATION. 


229 


Proposition  XVII.    Problem. 

441.  To  compute  the  ratio  of  the  circumference  of  a 
circle  to  its  diameter  approximately. 


Let  C  be  the  circumference,  and  R  the  radius. 
To  find  the  nume)i,cal  valine  ofir. 

27ri?=^Hy^ 
Therefore  when  -^^^Pl^^  C. 
We  make  the  following  computatrons  by  the  use  of  the 
formula  obtained  in  the  last  proposition,  when  i?  =  1,  and 
AB  =  1  (a  side  of  a  regular  hexagon). 


No. 
Bides. 


Fonn  of  Compatatlon. 


V2— V4^ 


Length  of  Side.       Length  of  Perimeter 


12  c,  =  V2-V4-P  0.51763809  6.21165708 

24  C2  =  ^^B^^^^^^^    0.26105238  6.26525722 

48  cs  -  V2  ^4  -  (0.2610523^,  -^.13080626  6.27870041 

96  c,=  V2-  V4-(0.13080626y.^^p.06533817  6.28206396 

192  C5  =  V2 -  V4 -  (0.0654381^    0.03272346  6.28290510 

0.01636228  6.28311544 
0.00818121 


6.28316941 


384  ^6  =  ^2 -V4- (0.03272346)2 

768  c,=  V2-V4-(0.01636228y 

Hence  we  may  consider  6.28317  as  approximately  the  cir- 
cumference of  a  0  whose  radius  is  unity. 

Therefore  7r--J(6.28317)  =  3.14159  nearly.       q.e.f. 

442.   Scholium.   In  practice,  we  generally  take 

1 


3.1416, 


0.31831. 


230 


PLANE    GEOMETRY.  —  BOOK    V, 


Maxima  and  Minima. — Supplementary. 

443.  Among  magnitudes  of  the  same  kind,  that  which  is 
greatest  is  the  maxiTnum,  and  that  which  is  smallest  is  the 
minimum. 

Thus  the  diameter  of  a  circle  is  the  maximum  among  all 
inscribed  straight  lines  ;  and  a  perpendicular  is  the  minimum 
among  all  straight  lines  drawn  from  a  point  to  a  given  line. 

444.  Isoperimetiic  figures  are  figures  which  have  equal 
perimeters. 

Proposition  XVIII.     Theorem. 

445.  Of  all  triangles  having  two  given  sides,  that 

in  which  these  sides  include  a  right  angle  is  the 

maxljnum. 

A 
E 


Let  the  triangles  ABC  and  EEC  have  the  sides  AB 
and  BC  equal  respectively  to  EB  and  BC ;  and  let  the 
angle  ABC  be  a  right  angle. 

To  prove  A  ABO  >  A  EEC. 

Proof.  From  E  let  fall  the  JL  ED. 

The  A  ABC ?iwdi  EBC,  having  the  same  base  BC,  are  to 

each  other  as  their  altitudes  AB  and  ED.  §  370 

Now                                EB  >  ED.  §  114 

By  hypothesis,  EB  =  AB. 

.'.  AB  >  ED. 

:.AABC>AEBQ.  q.e.d. 


[AXIMA    AND    MINIMA. 


231 


Proposition  XIX.     Theorem. 

446.   Of  all  triangles  having  the  same  base  and  equal 
perimeters,  the  isosceles  triangle  is  the  maximum. 


Let  the  ^  ACB  and  ADB  have  equal  perimeters,  and 
let  the  A  ACB  be  isosceles. 

To  prove  AACB>AA  DB. 

Proof.    Produce  AC  io  II,  making  CII=  AC,  and  join  HB. 
ABII'm  a  right  angle,  for  it  will  be  inscribed  in  the  semi- 
circle whose  centre  is  C,  and  radius  CA. 

Produce  HB,  and  take  I)F=  DB. 

Draw  C^and  DM  II  to  AB,  and  join  AP. 

Now  AII=  AC-\-  CB  -  AJ)^  DB  =  AD  +  DP. 

'^wt  AD -^  DP>AP,  hence  AII>  AP. 

Therefore  HB  >  BP.  §  120 

But  KB  =  ^  HB  and  3fB  =iBP.  §  121 

Hence  KB  >  MB. 
By  ^  180,  KB-^CE  and  MB  =  DF,  the  altitudes  of  the 
A  ACB  &,udi  ADB. 

Therefore  A  ABC>  A  ADB.  §  370 

Q.  E.  D 


232  PLANE  GEOMETRY.  —  BOOK  V. 


Proposition  XX.     Theorem. 

447.  Of  all  polygons  with  sides  all  given  hut  one, 
the  majcimuryb  can  he  inscribed  in  a  semicircle  which 
has  the  undetermined  side  for  its  diameter. 

c 


Let  ABODE  be  the  maximum  of  polygons  with  sides 
AD,  BC\  CD,  DE,  and  the  extremities  A  and  E  on  the 
straight  line  MN. 

To  prove  ABCDE  can  he  inscribed  in  a  semicircle. 

Proof.    From  any  vertex,  as  (7,  draw  QA  and  CE. 

The  A  ACE  must  be  the  maximum  of  all  A  having  the 
given  sides  CA  and  CE',  otherwise,  by  increasing  or  diminish- 
ing the  /.  ACE,  keeping  the  sides  CA  and  CE  unchanged,  but 
sliding  the  extremities  A  and  E  along  the  line  MN,  we  can 
increase  the  A  AGE,  while  the  rest  of  the  polygon  will  remain 
unchanged,  and  therefore  increase  the  polygon. 

But  this  is  contrary  to  the  hypothesis  that  the  polygon  is 
the  maximum  polygon. 

Hence  the  A  ACE  ^ffii\l  the  given  sides  CA  and  CE  is  the 
maximum. 

Therefore  the  Z.  ACE  i^  2^  right  angle,  §  445 

(the  maximum  of  A  having  two  given  sides  is  the  A  with  the  two  given  sides 
including  a  rt,  Z). 

Therefore  Clies  on  the  semi-circumference.         §  264 

Hence  every  vertex  lies  on  the  circumference ;  that  is,  the 

maximum  polygon  can  be  inscribed  in  a  semicircle  having  the 

undetermined  side  for  a  diameter.  a  e  o. 


MAXIMA    AND    MINIMA. 


233 


Proposition  XXI.     Theorem. 

448.  Of  all  -polygons  with  given  sides,  that  which 
van  be  inscribed  in  a  circle  is  the  maximum. 


Let  ABODE  be  a  polygon  inscribed  in  a  circle,  and 
A'B'C'D'E'  be  a  polygon,  equilateral  with  respect  to 
ABCDEf  which  cannot  be  inscribed  in  a  circle. 

To  prove    ABODE  greater  than  A'B'C'D'E^ 

Proof.  Draw  the  diameter  AIT. 

Join  (7^  and  BH. 
Upon  CD'  {=  CB)  construct  the  A  C'IBB'  -  A  CHB, 

and  draw  A' IP. 
Now  ABCH>  A'B'C'H\  §  447 

and  AEBH>  A'E'B'W, 

{of  all  polygons  with  sides  all  given  hut  one,  the  maximum  can  he  inscribed 
in  a  semicircle  having  me  undetermined  side  for  its  diameter). 

Add  these  two  inequalities,  then 

ABCHBE  >  A'B'C'BT'B'E'. 

Take  away  from  the  two  figures  the  equal  A  CB^B  and  C^H^BK 

Then  ABCBE  >  A'B'C'B'E'.  cle.d. 


234 


PLANE   GEOMETRY.  —  BOOK    V. 


Proposition  XXII.     Theorem. 

449.   Of  isoperimetric  polygons  of  the  same  number 
of  sides,  the  maximum  is  equilateral. 

K 


Let  ABCD,  etc.,   be    the   maximum,   of  isoperimetric 
polygons  of  any  given  number  of  sides. 

To  prove  AJB,  BC,  CD,  etc.,  equal. 

Proof.  Draw  AC. 

The  A  ABC  must  be  the  maximiim  of  all  the  A  which  are 
forraed  upon  J. C  with  a  perimeter  equal  to  that  of  A  ABC. 

Otherwise,    a   greater  A  AKC  could   be   substituted   for 
A  ABC,  without  changing  the  perimeter  of  the  polygon. 

But  this  is  inconsistent  with  the  hypothesis  that  the  poly- 
gon ABCD,  etc.,  is  the  maximum  polygon. 

.-.  the  A  ABCh  isosceles,  §  446 

{of  all  A  having  the  same  base  and  equal  perimeters,  the  isosceles  A  is  the 
maximum). 

In  like  manner  it  may  be  proved  that  BC=  CD,  etc.  q.e.  d. 

450.    Cor.    The  maxiTnum  of  isoperimetric  polygons  of  the 

same  number  of  sides  is  a  regular  polygon.     ■ 

For,  it  is  equilateral,  §  449 

{the  maximum  of  isoperimetric  polygons  of  the  same  number  of  sides  is 
equilateral). 

Also  it  can  be  inscribed  in  a  circle,  §  448 

{the  maximum  of  all  polygons  formed  of  given  sides  can  be  inscribed  in  a  O). 

That  is,  it  is  equilateral  and  equiangular, 

and  therefore  regular.  §  395 

Q.  E.  D. 


MAXIMA  AND   MINIMA. 


235 


Proposition  XXIII.    Theorem. 

451.  Of  isoperimetric  regular  polygons,  that  which 
has  the  greatest  number  of  sides  is  the  Tnaximunv, 
O 


A         D         B 

Let  Q  be  a  regular  polygon  of  three  sides,  and  Q' 
a  regular  polygon  of  four  sides,  and  let  the  two  poly- 
gons have  equal  perimeters. 

To  prove  Q'  greate)'  than  Q. 

Proof.       Draw  CD  from  Cto  any  point  in  AB. 

Invert  the  A  CDA  and  place  it  in  the  position  DCE,  let- 
ting D  fall  at  (7,  Cat  Z>,  and  A  at  E. 

The  polygon  DBCE  is  an  irregular  polygon  of  four  sides, 
which  by  construction  has  the  same  perimeter  as  Q ,  and  the 
same  area  as  Q. 

Then  the  irregular  polygon  DBCE  of  four  sides  is  less  than 
the  regular  isoperimetric  polygon  Q  of  four  sides.  §  450 

In  like  manner  it  may  be  shown  that  Q  is  less  than  a  regular 
isoperimetric  polygon  of  five  sides,  and  so  on.  q.  e,  d. 

452.  Cor.  The  area  of  a  circle  is  greater  than  the  area  of 
any  polygon  of  equal  perimeter. 

\382.  Of  all  equivalent  parallelograms  having  equal  bases,  the  rec- 
tangle has  the  least  perimeter. 

'^SSS.   Of  all  rectangles  of  a   given  area,  the  square  has  the  least 
perimeter. 

^384.   Of  all  triangles  upon  the  same  base,  and  having  the  same  alti- 
tude, the  isosceles  has  the  least  perimeter. 

^^85.  To  divide  a  straight  line  into  two  parts  such  that  their  product 
shall  be  a  maximum. 


^. 


236 


PLANE   GEOMETRY. —  BOOK   V. 


Proposition  XXIV.     Theorem. 

453.  Of  regular  -poly ions  having  a  given  area,  that 
which  has  the  greatest  number  of  sides  has  the  least 
perimeter. 


Q' 


Let  Q  and  Q'  be  regular  polygons  having  the  same 
area,  and  let  Q'  have  the  greater  number  of  sides. 

To  prove  the  perimeter  of  Q  greater  than  the  perimeter  of  Q\ 

Proof.    Let  Q"  be  a  regular  polygon  having  the  same  perim- 
eter as  Q\  and  the  same  number  of  sides  as  Q. 

Then  Q  >  Q\  §  461 

{of  isoperimetric  regular  polygons,  that  which  has  the  greoAest  number  of 
sides  is  the  maximum). 

But  Q=Q'. 

.-.  Q  >  Q". 

.'.  the  perimeter  of  Q>  the  perimeter  of  Q". 
But  the  perimeter  of  Q'  =  the  perimeter  of  Q".         Cons. 
.•.  the  perimeter  of  §  >  that  of  Q'. 


454.    Cor.   The  circumference  of  a  circle  is   less   than  the 
perimeter  of  any  polygon  of  equal  area. 


386.  To  inscribe  in  a  semicircle  a  rectangle  having  a  given  area  ; 
a  rectangle  having  the  maximum  area. 

^S»387.   To  find  a  point  in  a  semi-circumference  such  that  the  sum  of  its 
'  distances  from  the  extremities  of  the  diameter  shall  be  a  maximum. 


EXERCISES.  237 


Theorems. 

388.  The  side  of  a  circumscribed  equilateral  triangle  is  equal  to  twice 
the  side  of  the  similar  inscribed  triangle.    Find  the  ratio  of  their  areas. 

389.  The  apothem  of  an  inscribed  equilateral  triangle  is  equal  to  half 
the  radius  of  the  circle. 

390.  The  apothem  of  an  inscribed  regular  hexagon  is  equal  to  half 
the  side  of  the  inscribed  equilateral  triangle. 

391.  The  area  of  an  inscribed  regular  hexagon  is  equal  to  three- 
fourths  of  that  of  the  circumscribed  regular  hexagon. 

392.  The  area  of  an  inscribed  regular  hexagon  is  a  mean  proportional 
bulween  the  areas  of  the  inscribed  and  the  circumscribed  equilateral 
triangles. 

393.  The  area  of  an  inscribed  regular  octagon  is  equal  to  that  of  a 
rectangle  whose  sides  are  equal  to  the  sides  of  the  inscribed  and  the  cir- 
cumscribed squares. 

394.  The  area  of  an  inscribed  regular  dodecagon  is  equal  to  three 
times  the  square  of  the  radius. 

395.  Every  equilateral  polygon  circumscribed  about  a  circle  is  regu- 
lar if  it  has  an  odd  number  of  sides. 

396.  Every  equiangular  polygon  inscribed  in  a  circle  is  regular  if  it 
has  an  odd  number  of  sides. 

397.  Every  equiangular  polygon  circumscribed  about  a  circle  is 
regular. 

398.  Upon  the  six  sides  of  a  regular  hexagon  squares  are  constructed 
outwardly.  J  rove  that  the  exterior  vertices  of  these  squares  are  the  ver- 
tices of  a  regular  dodecagon. 

399.  The  alternate  vertices  of  a  regular  hexagon  are  joined  by  straight 
lines.  Vrove  that  another  regular  hexagon  is  thereby  formed."  Find  the 
ratio  of  the  areas  of  the  two  hexagons. 

400.  The  radius  of  an  inscribed  regular  polygon  is  the  mean  propor- 
tional between  its  apothem  and  the  radius  of  the  similar  circumscribed 
regular  polygon. 

401.  The  area  of  a  circular  ring  is  equal  to  that  of  a  circle  whose 
diameter  is  a  chord  of  the  outer  circle  and  a  tangent  to  the  inner  circle. 

402.  The  square  of  the  side  of  an  inscribed  regular  pentagon  is  equal 
to  the  sum  of  the  squares  of  the  radius  of  the  circle  and  the  side  of  the 
inscribed  regular  decagon. 


238  PLANE    GEOMETRY.  —  BOOK    V. 

If  R  denotes  the  radius  of  a  circle,  and  a  one  side  of  a  regular  inscribed 
polygon,  show  that : 

R 


403.    In  a  regular  pentagon,  a  =       -y  iq  —  2"\/5. 


404.    In  a  regular  octagon,     cb  =  R-^ 2—  \/2. 


405.  In  a  regular  dodecagon,  a  =  R  Vz-^Vl. 

406.  If  on  the  legs  of  a  right  triangle,  as  diameters,  semicircles  are 
described  external  to  the  triangle,  and  from  the  whole  figure  a  semicircle 
on  the  hypotenuse  is  subtracted,  the  remainder  is  equivalent  to  the  given 
triangle. 

Numerical  Exercises. 

407.  The  radius  of  a  circle  ==  r.  Find  one  side  of  the  circumscribed 
equilateral  triangle. 

408.  The  radius  of  a  circle  =  r.  Find  one  side  of  the  circumscribed 
regular  hexagon. 

409.  If  the  radius  of  a  circle  is  r,  and  the  side  of  an  inscribed  regular 
polygon  is  a,  show  that  the  side  of  the  similar  circumscribed  regular 

)olygon  is  equal  to  2ar 

V'4r^-a2 

"N410.  The  radius  of  a  circle  =  r.  Prove  that  the  area  of  the  inscribed 
regular  octagon  is  equal  to  2r'^V2. 

■^411.  The  sides  of  three  regular  octagons  are  3  feet,  4  feet,  and  5  feet, 
respectively.  Find  the  side  of  a  regular  octagon  equal  in  area  to  the 
sum  of  the  areas  of  the  three  given  octagons.— y^-^ 

M12.  What  is  the  width  of  the  ring  between  two  concentric  circum- 
ferences whose  lengths  are  440  feet  and  330  feet  ?  0»  t)  -f 

^413.  Find  the  angle  subtended  at  the  centre  by  an  arc  6  feet  5  inches 
long,  if  the  radius  of  the  circle  is  8  feet  2  inches.    ^  ^"^  -f- 

>414.  Find  the'  angle  subtended  at  the  centre  of  a  circle  by  an  arc 
whose  length  is  equal  to  the  radius  of  the  circle.    6'"'^«  ^^ ' 

•415.   What  is  the  length  of  the  arc  subtended  by  one  side  of  a  regular/  «|L 
dodecagon  inscribed  in  a  circle  whose  radius  is  14  feet?  -v-.'V^  "i^i"   §^ 

v416.  Find  the  side  of  a  square  equivalent  to  a  circle  whose  radius  is 
86  feet     f<f.l^jt>i 


EXERCISES.  239 

^417.  Find  the  area  of  a  circle  msc^Pibed  in  a  square  containing  196 
square  feet.  /  3'  ^  ,9   ^  ^  V  ^Jt  f^  * 

M:18.  The  diameter  of  a  circular  grass  plot  is  28  feet.  ,  Eind  the  diam- 
eter of  a  circular  plot  just  twice  as  large.  _j  ^«  ^  "*    f-^ 

*419.  Find  the  side  of  the  largest  square  that  can  be  cut  out  of  a  cir- 
cular piece  of  wood  whose  radius  is  1  foot  8  inches.    ^    /  "^  /      • 

M20.  The  radius  of  a  circle  is  3  feet.  What  is  thejadius  of  a  circle  25 
times  as  lp,rge  ?   |  as  large  ?   -jJjy  as  large  ?  -,      O'T* ^ 

•-421.  The  radius  of  a  circle  is  9  feet.  What  are  the  radii  of  the  con- 
centric circumferences  that  will  divide  the  circle  into  three  equivalent 
parts?    ^.1-     c^y^A^      "^i^" 

^422.  The  chord  of  half  an  arc  is  12  feet,  and  the  radius  of  the  circle  is 
18  feet.     Find  the  height  of  the  arc. 

423.  The  chord  of  an  arc  is  24  inches,  and  the  height  of  the  arc  is.-9 
inches.     Find  the  diameter  of  the  circle. 

424.  Find  the  area  of  a  sector,  if  the  radius  of  the  circle  is  28  feet, 
and  the  angle  at  the  centre  22J°. 

425.  The  radius  of  a  circle  =  r.  Find  the  area  of  the  segment  sub- 
tended by  one  side  of  the  inscribed  regular  hexagon. 

426.  Three  equal  circles  are  described,  each  touching  the  other  two. 
If  the  common  radius  is  r,  find  the  area  contained  between  the  circles. 

Problems. 

To  circumscribe  about  a  given  circle  : 

427.  An  equilateral  triangle.  429.   A  regular  hexagon. 

428.  A  square.  430.   A  regular  octagon. 

431.  Totdraw  through  a  given  point  a  line  so  that  it  shall  divide  a 
given  circumference  into  two  parts  having  the  ratio  3 :  7. 

432.  To  construct  a  circumference  equal  to  the  sum  of  two  given 
circumferences. 

433.  To  construct  a  circle  equivalent  to  the  sum  of  two  given  circles. 

434.  To  construct  a  circle  equivalent  to  three  times  a  given  circle. 

435.  To  construct  a  circle  equivalent  to  three-fourths  of  a  given  circle. 
To  divide  a  given  circle  by  a  concentric  circumference : 

436.  Into  two  equivalent  parts.        437.   Into  five  equivalent  parts. 


240  plane  geometry.  —  book  v. 

Miscellaneous  Exercises. 
Theorems. 

438.  The  line  joining  the  feet  of  the  perpendiculars  dropped  from  the 
extrertiities  of  the  base  of  an  isosceles  triangle  to  the  opposite  sides  is 
parallel  to  the  base. 

439.  If  AD  bisect  the  angle  J.  of  a  triangle  ABC,  and  BD  bisect  the 
exterior  angle  CBF,  then  angle  ADB  equals  one-half  angle  ACB. 

440.  The  sum  of  the  acute  angles  at  the  vertices  of  a  pentagram  (five- 
pointed  star)  is  equal  to  two  right  angles. 

441.  The  bisectors  of  the  angles  of  a  parallelogram  form  a  rectangle. 

442.  The  altitudes  AD,  BE,  CF  of  the  triangle  ABC  bisect  the  angles 
of  the  triangle  DEF. 

Hint.  Circles  with  AB,  BC,  ^C  as  diameters  will  pass  through  E  and 
D,  E  and  F,  D  and  F,  respectively. 

443.  The  portions  of  any  straight  line  intercepted  between  the  cir- 
cumferences of  two  concentric  circles  are  equal. 

444.  Two  circles  are  tangent  internally  at  P,  and  a  chord  AB  of  the 
larger  circle  touches  the  smaller  circle  at  C.  Prove  that  PC  bisects  the 
angle  AFB. 

Hint.   Draw  a  common  tangent  at  P,  and  apply  §g  263,  269,  145, 

445.  The  diagonals  of  a  trapezoid  divide  each  other  into  segments 
which  are  proportional. 

446.  The  perpendiculars  from  two  vertices  of  a  triangle  upon  the 
opposite  sides  divide  each  other  into  segments  reciprocally  proportional. 

447.  If  through  a  point  P  in  the  circumference  of  a  circle  two  chords 
are  drawn,  the  chords  and  the  segments  between  P  and  a  chord  parallel 
to  the  tangent  at  P  are  reciprocally  proportional. 

448.  The  perpendicular  from  any  point  of  a  circumference  upon  a 
chord  is  a  mean  proportional  between  the  perpendiculars  from  the  same 
point  upon  the  tangents  drawn  at  the  extremities  of  the  chord. 

449.  In  an  isosceles  right  triangle  either  leg  is  a  mean  proportional 
between  the  hypotenuse  and  the  perpendicular  upon  it  from  the  vertex 
of  the  right  angle. 

450.  The  area  of  a  triangle  is  equal  to  half  the  product  of  its  perim- 
eter by  the  radius  of  the  inscribed  circle. 


MISCELLANEOUS   EXERCISES.  241 

451.  The  perimeter  of  a  triangle  is  to  one  side  as  the  perpendicular 
from  the  opposite  vertex  is  to  the  radius  of  the  inscribed  circle. 

452.  The  sum  of  the  perpendiculars  from  any  point  within  a  convex 
equilateral  polygon  upon  the  sides  is  constant. 

453.  A  diameter  of  a  circle  is  divided  into  any  two  parts,  and  upon 
these  parts  as  diameters  semi-circumferences  are  described  on  opposite 
sides  of  the  given  diameter.  Prove  that  the  sum  of  their  lengths  is  equal 
to  the  semi-circumference  of  the  given  circle,  and  that  they  divide  the 
circle  into  two  parts  whose  areas  have  the  same  ratio  as  the  two  parts 
into  which  the  diameter  is  divided. 

454.  Lines  drawn  from  one  vertex  of  a  parallelogram  to  the  middle 
points  of  the  opposite  sides  trisect  one  of  the  diagonals. 

455.  If  two  circles  intersect  in  the  points  A  and  B,  and  through  A 
any  secant  CAD  is  drawn  limited  by  the  circumferences  at  C  and  D,  the 
straight  lines  BC,  BD,  are  to  each  other  as  the  diameters  of  the  circles. 

456.  If  three  straight  lines  AA^,  BB\  CC^,  drawn  from  the  vertices 
of  a  triangle  ABC  to  the  opposite  sides,  pass  through  a  common  point  0 
within  the  triangle,  then 

OA'    OB'    oa  ^. 

AA'     BB'      CC      ' 

457.  Two  diagonals  of  a  regular  pentagon,  not  drawn  from  a  common 
vertex,  divide  each  other  in  extreme  and  mean  ratio. 

Loci. 

458.  Find  the  locus  of  a  point  P  whose  distances  from  two  given 
points  A  and  B  are  in  a  given  ratio  (m :  n). 

459.  OP  is  any  straight  line  drawn  from  a  fixed  point  0  to  the  cir- 
cumference of  a  fixed  circle ;  in  OP  a  point  Q  is  taken  such  that  OQ:  OP 
is  constant.    Find  the  locus  of  Q. 

460.  From  a  fixed  point  A  a  straight  line  AB  is  drawn  to  any  point 
in  a  given  straight  line  CD,  and  then  divided  at  P  in  a  given  ratio 
(m  :  n).     Find  the  locus  of  the  point  P. 

461.  Find  the  locus  of  a  point  whose  distances  from  two  given  straight 
lines  are  in  a  given  ratio.     (The  locus  consists  of  two  straight  lines.) 

462.  Find  the  locus  of  a  point  the  sum  of  whose  distances  from  two 
given  straight  lines  is  equal  to  a  given  length  k.    (See  Ex.  73.) 


242  PLANE   GEOMETRY. BOOK    V. 


Problems. 

463.  Given  the  perimeters  of  a  regular  inscribed  and  a  similar  circum- 
scribed polygon,  to  compute  the  perimeters  of  the  regular  inscribed  and 
circumscribed  polygons  of  double  the  number  of  sides. 

464.  To  draw  a  tangent  to  a  given  circle  such  that  the  segment  inter- 
cepted between  the  point  of  contact  and  a  given  straight  line  shall  have 
a  given  length. 

465.  To  draw  a  straight  line  equidistant  from  three  given  points. 

466.  To  inscribe  a  straight  line  of  given  length  between  two  given 
circumferences  and  parallel  to  a  given  straight  line.     (See  Ex.  137.) 

467.  To  draw  through  a  given  point  a  straight  line  so  that  its  dis- 
tances from  two  other  given  points  shall  be  in  a  given  ratio  (m :  n). 

Hint.    Divide  the  line  joining  the  two  other  points  in  the  given  ratio. 

468.  Construct  a  square  equivalent  to  the  sum  of  a  given  triangle 
and  a  given  parallelogram. 

469.  Construct  a  rectangle  having  the  difference  of  its  base  and 
altitude  equal  to  a  given  line,  and  its  area  equivalent  to  the  sum  of  a 
given  triangle  and  a  given  pentagon. 

470.  Construct  a  pentagon  similar  to  a  given  pentagon  and  equiva- 
lent to  a  given  trapezoid. 

471.  To  find  a  point  whose  distances  from  three  given  straight  lines 
shall  be  as  the  numbers  m,  n,  and  p.     (See  Ex.  461.) 

472.  Given  two  circles  intersecting  at  the  point  A.  To  draw  through 
A  a  secant  B AC  such  that  AB  shall  be  to  ^Cin  a  given  ratio  (m  :  n). 

Hint.    Divide  the  line  of  centres  in  the  given  ratio. 

473.  To  construct  a  triangle,  given  its  angles  and  its  area. 

474.  To  construct  an  equilateral  triangle  having  a  given  area. 

475.  To  divide  q,  given  triangle  into  two  equal  parts  by  a  line  drawn 
parallel  to  one  of  the  sides. 

476.  Given  three  points  A,  ^,  C.  To  find  a  fourth  point  P  such  that 
the  areas  of  the  triangles  AFB,  APC,  BFC,  shall  be  equal.    . 

477.  To  construct  a  triangle,  given  its  base,  the  ratio  of  the  other 
sides,  and  the  angle  included  by  them. 

478.  To  divide  a  given  circle  into  any  number  of  equivalent  parts  by 
concentric  circumferences. 

479.  In  a  given  equilateral  triangle,  to  inscribe  three  equal  circles 
tangent  to  each  other  and  to  the  sides  of  the  triangle. 


SOLID  GEOMETRY. 

BOOK   VI. 
LINES    AND    PLANES    IN    SPACE. 


Definitions. 

455.  A  plane  has  already  been  defined  as  a  surface  such 
that  a  straight  line  joining  any  two  points  in  it  lies  wholly 
in  the  surface. 

A  plane  is  considered  to  be  indefinite  in  extent,  so  that 
however  far  the  straight  line  is  produced,  all  its  points  lie  in 
the  plane  ;  but  a  plane  is  usually  represented  by  a  quadrilat- 
eral supposed  to  lie  in  the  plane. 

456.  A  plane  is  said  to  be  determined  by  lines  or  points,  if 
no  other  plane  can  contain  these  lines  or  points  without  being 
coincident  with  that  plane. 

457.  A  plane  can  be  n\ade  to  turn  about  any  straight  line 
in  it  as  an  axis,  and  be  made  to 

assume  as  many  different  posi- 
tions as  we  choose.     Hence  it  is 

evident  that  a  plane  is  not  deter-    -^^^~~^ J 

mined  by  a  straight  line.  y      '  / 

In  making  a  complete  revolu- 
tion about  a  straight  line  as  an 
axis  the  plane  passes  successively  through  all  points  of  space. 

458.  A  plane  is  determined  by  a  straight  line  and  a  point 
without  thai  line. 


I 


N 


244  SOLID    GEOMETEY.  —  BOOK    VI. 

If  a  plane  containing  the  straight  line  AB  revolve  about 
this  line  as  an  axis  until  it  con-  t^j 

tains  the  point  C,  the  plane  is 
determined.     For  if  the  plane      y^      A^ 

revolve  either  way  about  the     ''^ 

line  AB  as  an  axis,  it  will  cease  to  contain  the  point  C 

459.  Three  points  not  in  a  straight  line  determine  a  plane. 
For,  by  joining  any  two  of  the  points  we  have  a  straight 

line  and  a  point  without  it,  and  these  determine  a  plane.  §  458 

460.  Two  intersecting  straight  lines  determine  a  plane. 

For,  a  plane  containing  one  of  these  straight  lines  and  any 
point  of  the  other  line  in  addition  to  the  point  of  intersection 
is  determined.  §  458 

461.  Two  parallel  straight  lines-  determine  a  plane. 
For,  two  parallel  straight  lines  lie  in 

iiiSr  same  ;£lane,  and  a  plane  containing 
either  of  these  parallels  and  any  point 
in  the  other  is  determined.  §  458 

462.  A  straight  line  is  perpendicular  to  a  plane  if  it  is  per- 
pendicular to  every  straight  line  of  the  plane  drawn  through' 
its  foot ;  that  is,  through  the  point  where  it  meets  the  plane. 

In  this  case  the  plane  is  perpendicular  to  the  line. 

463.  A  line  is  oblique  to  a  plane  if  it  is  not  perpendicular 
to  all  straight  lines  drawn  in  the  plane  through  its  foot. 

464.  The  distance  from  a  point  to  a  plane  is  the  perpen- 
dicular distance  from  the  point  to  the  plane. 

465.  A  line  is  parallel  to  a  plane  if  it  cannot  meet  the  plane 
however  far  both  are  produced. 

In  this  case  the  plane  is  parallel  to  the  line. 

466.  Two  plpLnes  are  parallel  if  they  cannot  meet  however 
far  they  are  produced. 


LINES   AND    PLANES   IN   SPACE. 


245 


467.  The  projection  of  a  point  on  a  plane  is  the  foot  of  the 
perpendicular   from   the   point  to  the  a.  B 
plane. 

468.  The  projection  of  a  line  on  a 
plane  is  the  locus  of  the  projections  of 
all  its  points. 

469.  The  angle  which  a  line  makes 
with  a  plane  is  the  angle  which  it  makes  with  its  projection 
on  the  plane. 

470.  The  intersection  of  two  planes  is  the  locus  of  all  the 
points  common  to  the  two  planes. 

Proposition  I.     Theorem. 

471.  If  two  planes  cut  each  other,  their  intersection 
is  a  straight  line. 


Let  MN  and  PQ  be  two  planes  which  cut  one  another. 

To  prove  their  intersection  a  straight  line. 

Proof.  Let  A  and  B  be  two  points  common  to  the  two  planes. 
Draw  a  straight  line  through  the  points  A  and  B. 

Since  the  points  A  and  B  are  common  to  the  two  planes, 
this  straight  line  lies  in  both  planes.  §  455 

No  point  out  of  this  line  can  be  in  both  planes  ;  for  only  one 
plane  can  contain  a  straight  line  and  a  point  without  the  line. 

Therefore  the  straight  line  through  A  and  B  is  the  locus  of 
all  the  points  common  to  the  two  planes,  and  is  consequently 
the  intersection  of  the  planes  (§  470).  q.  e.d. 


246 


SOLID    GEOMETEY. 


BOOK    VI. 


Perpendicular  Lines  and  Planes. 
Proposition  II.     Theorem. 

472.  If  a  straight  line  is  perpendicular  to  each  of 
two  other  straight  lines  at  their  point  of  intersection, 
it  is  perpendicular  to  the  plane  of  the  two  lines. 


\l' 


Let  AB  be  perpendicular  to  BG  and  BD  at  B. 

To  prove  A B  perpendicular  to  the  plane  MN  of  these  lines. 
Proof.    Through  B  draw  in  MJV  any  other  straight  line  BE, 
and  draw  CD  cutting  £C,  BE,  BD,  at  O,  E,  and  D. 

Prolong  ^^  to  i^,  making  BF=j=  AB,  and  join  J.  and  i^to 
each  of  the  points  C,  E,  and  D. 

Since  BO  and  BD  are  each  J_  to  ^i^at  its  middle  point, 

•    AC=='FC^rid.AD  =  FD.  §122 

.-.  AACD  =  A  FCD  (§  160),  and  hence  AACD^Z  FCD. 
Now  in  the  A  ACE  and  FCE 

AC=  EC,  CE=  CE,  and  Z.  ACE==  A  FCE. 
.-.  A  ACE^-  A  FCE  {%  150),  and  hence  AE=  FE. 

.'.  BE  is  ±  to  AF  Sit  B.  §123 

Hence  ^i?  is  _L  to  BE,  any,  that  is,  every,  straight  line 

drawn  in  MN  through  B,  and  therefore  is  ±  to  MN.      ^  462 

Q  E.  D. 


PERPENDICULAR    LINES   AND    PLANES. 


247 


Proposition  III.     Theorem. 

473.  Every  perpendicular  that  can  he  drawn  to  a 
straight  line  at  a  given  point  lies  in  a  plane  perpen- 
dicular to  the  line  at  tlve  given  point, 

A 


Let  the  plane  MN  be  perpendicular  to  AB  at  B. 

To  prove  that  BE,,  any  perpendicidar  to  AB  at  B,  lies  in 
the  plane  MN.- 

Proof.  Let  the  plane  containing  ABi^  and  BE  intersect  MN 
in  the  line  BE' ;  then  AB  is  ±  to  J^\      .  §  462 

Since  in  the-p^ttrne  ABE  only  one  JL  can  be  drawn  to  AB  at- 
B  (§  89),  BE  and  BE^  coincide,  and  BE  lies  in  MN 

Hence  every  ±  to  AB  at  B  lies  in  the  plane  MN. 

•^  Q.  E.  D. 

474.  Cor.  1.  At  a  given  point'  m  a  straight  line  one  plane 
perpendicular  to  the  line  can  he  draivn,  and  only  one. 

475.  Cor,  2.   Through  a  given  point  without  a  straight  line, 
one  plane  can  he  drawn  perpendicular  to 
the  line,  and  only  one. 

Let  AC  hQ  the  line,  and  0  the  point 
without  it.  In  the  plane  OCA  draw  OC 
J_  to  AC,  .and  in  another  plane  contain- 
ing ^ a  draw  CD  A.  to  AC  2X  C.  Then 
CO  and  CD  determine  a  plane  l.io  AC  x 

Every  plane  JL  to  AC  B,nd.  passing  through  0  cuts  the  plane 
OCA  in  a  line  ±  to  ^C  and  containing  0.  This  ±  coincides, 
then,  with  OC,  and  every  such  plane  is  ±  to  ^C  at  0.  But 
only  one  plane  can  be  ±  to  ^(7at  C(§  474).  Hence  only  one 
plane  can  be  drawn  from  0  X  to  ^Cat  0. 


248 


SOLID    GEOMETET.  —  BOOK    VI. 


Proposition  IV.     Theorem. 

476.  Through  a  given  point  one  perpendicular  can 
he  drawn  to  a  given  plane,  and  only  one, 

Pr 


K 

H 

\ 

'  1 

"/ 

^ 

/ 

$ 

/ 

'B 

f 

/ 

-^ 

M 

Fig.  1.  

Fig.  2.  Q 

Case  I.   When  the  given  point  is  in  the  given  plane. 

Let  A  be  the  given  point  in  the  plane  MN  (Fig.  1). 

To  prove  that  one  perpendicular  can  be  erected  to  the  plane 
MN  at  A,  and  only  one. 

Proof.  Draw  in  JfiVany  line  -SC'througli  A,  and  pass  through 
A  a  plane  DEHK  JL  to  BC,  and  cutting  MN'm  DE. 

At  A  erect  in  the  plane  DEHK  a  line  AF  X  to  DE. 

The  line  BQ,  being  J_  to  the  plane  DEHKhj  construction, 
is  J_  to  ^i^  which  passes  through  its  foot  in  the  plane.    §  462 

That  is,  AF  is  JL  to  BC)  and  as  it  is  JL  to  DE  by  con- 
struction, it  is  X  to  the  plane  MN.  §  472 

Moreover,  every  other  line  AG  drawn  from  A,  is  oblique  to 
MN.  For  ^i^and  AO  intersecting  in  A  determine  a  plane 
DEHK,  which  cuts  MN'm  the  straight  line  DE;  and  as  AF 
is  X  to  MN,  it  is  X  to  DE  (§  462) ;  hence  ^G^  is  oblique  to 
DE  (§  89),  and  therefore  to  Mn[^  463). 

Therefore  ^jPis  the  only  X  to  ifiV  at  the  point  A. 


PERPENDICULAR    LINES   AND    PLANES.  249 

Case  II.   When  the  given  point  is  without  the  given  'plane. 

Let  A  be  the  given,  point,  and  MN  the  plane. 

To  prove  that  one  perpendicular  can  be  drawn  from  A  to 
MN,  and  only  one. 

Proof.  Draw  in  3£N  any  line  HK,  and  pass  through  A  a 
plane  FQ  L  to  HK,  cutting  MN  in  FO,  and  HK  in  C. 

Let  fall  from  A,  in  the  plane  FQ,  ^  A.  AB  upon  FO. 

Draw  in  the  plane  MN  B.ny  other  line  BF  from  B. 

Prolong  AB  to  F,  making  BF=  AB, 

and  join  A  and  F  to  each  of  the  points  Cand  F. 

Since  i)C  is  _L  to  FQ  by  construction,  and  CA  and  CF  lie 
in  FQ,  the  A  FCA  and  DC^  are  right  angles.  §  462 

In  the  rt.  A  FCA  and  FCF, 

FCis  common,  and  CA  =  CF.  §  122 

.-.  A  FCA  =  A  FCE  (^  151),  and  hence  FA  =  FF. 

.-.  BF  is  ±  to  AF&tB.  §  123 

That  is,  AB  is  JL  to  BF,  any  straight  line  drawn  in  MN 
through  its  foot,  and  therefore  _L  to  MN 

Moreover,  every  other  straight  line  AI,  drawn  from  A,  is 
oblique  to  MN  For  the  lines  AB  and  ^/determine  a  plane 
FG  which  cuts  the  plane  MN  in  the  line  FG.  The  line  AB 
being  ±  to  the  plane  MN,  is  _L  to  FG  (§  462).    Therefore  AI 

oblique  to  FG,  and  consequently  to  MN  (^  463). 

Therefore  AB  is  the  only  ±  from  A  to  MN. 

Q.  E.  O. 

477.  Cor.  The  perpendicular  is  the  sho7'testUne  from  a  point 
to  a  plane,  for  it  is  the  shortest  line  from  the  point  to  any  straight 
line  of  the  plane  passing  through  its  foot  (§  114).  y 


[ 


250 


SOLID    GEOMETEY. 


BOOK    VI. 


Proposition  V.     Theorem. 

478.  Oblique  liiies  draivn  from  a  point  to  a  plane, 
and  meeting  the  plane  at  equal  distances  froin .  the 
foot  of  the  perpendicular,  are  equal;  and  of  two  ob- 
lique lines  meeting  the  plane  at  unequal  distances 
from  the  foot  of  the  perpendicular  the  more  remote 
is  the  greater. 


Let  AC  and  AD  cut  off  the  equal  distances  BC  and 
BD  from  the  foot  of  the  perpendicular  AB,  and  let  AD 
and  AE  cut  off  the  unequal  distances  BD  and  BE,  and 
BE  be  greater  than  BD. 

To  prove  AC=  AD,  and  AE  >  AD. 

Proof.  The  right  A  ABCd^ndi  ADD  have  AB  common,  and 
£0=  BD  by  hypothesis. 

Therefore  they  are  equal,  and  AC  =^  AD. 
The  right  A ^^^,  ^^C'have  AB  common,  and  BE>  BC. 
Therefore  AE>  ACi%  119),  and  hence  AE>  AD. 

Q.  E.  D, 

479.  Cor.  1.  Equal  oblique  lines  from  a  point  to  a  plane 
vieet  the  plane  at  equal  distances  from  the  foot  of  the  perpen- 
dicular ;  and  of  two  unequal  oblique  lines  the  greater  meets  the 
plane  at  the  greater  distance  from  the  foot  of  the  perpendicular. 

480.  Cor.  2.  The  locus  of  a  point  in  space  equidistant  from 
all  points  in  the  circumference  of  a  circle  is  a  straight  line  pass- 
ing through  the  centre  and  perpendicular  to  the  plane  of  the 
circle. 


PERPENDICULAR    LINES   AND    PLANES. 


251 


Proposition  VI.     Theorem. 

481.  If  from  the  foot  of  a  perpendicular  to  a  plane 
a  straight  line  is  drawn  at  right  angles  to  any  line 
in  the  plane,  the  line  drawn  from  its  intersection 
with  the  line  in  the  plane  to  any  point  of  t,%e  per- 
pendicular is  perpendicular  to  the  line  of  the  plane. 


M 

^N 

— 

_ 

'»  \,--'' 

■"^'  v 

B 

-•-... 

'•  r 

Let  AB-be  a  perpendicular  to  the  plane  MN,  BE  a 
perpendicular  from  B  to  any  line  CD  in  MN,  and  EA 
^  line  from  E  to  any  point  A  in  AB. 

To  prove  AE perpendiculo.r  to  CD. 

Proof.   Take  ^(7 and  ED  equal ;  draw  BC\  BD,  AC,  AD. 

Now  BC=  BD  (§  116),  and  AC=  AD  (§  478). 

:.AE\^l.\oCD,  §123 


482.  Cor.  The  locus  of  a  point  in  space  equidistant  from  the 
extremities  of  a  straight  line  is  the  plane  perpendicular  to  this 
line  at  its  middle  point. 

For,  if  the  plane  MN  is  ±  to  ^^  at  its 
middle  point  0,  and^any  point  C  in  this         ^^fc^ 
plane  is  joined  to  A,  0,  and  B,  CO  is  ±  to 
AB  ;  therefore  CA  and  CB  are  equal.  §  116  ^ 

Also,  since  all  the  Js  to  the  line  AB  at 
the  point  0  lie  in  the  plane  3/iV^(§473), 
any  point  D  without  the  plane  MN  cannot  lie  in  a  ±  to  AB 
at  0,  and  therefore  is  unequally  distant  from  A  and  B.  §  122 


252 


SOLID    GEOMETKY. 


BOOK    VI. 


Proposition  VII.     Theorem. 

483.  Two  straight  lines  perpendicular  to  the  same 
plane  are  parallel. 

/  A  C 


§462 
§481 
lie  in 
§473 


i^ 


Let  AB  and  CD  bs  perpendicular  to  MN. 

To  prove  AB  and  CD  parallel. 

Proof.    Let  A  be  any  point  in  AB-,  join  AD  and  BD,  and 
tlirough  D  draw  ^i^in  the  plane  UN  1.  to  BD. 
Then  CD  is  ±  to  ER 

Also,  AD  is  ±  to  EF. 

Therefore  CD,  AD,  and  BD,  being  _L  to  EF  at  D 
the  same  plane. 

Therefore  AB  and  CD  lie  in  the  same  plane  ;  and  since,  by 

hypothesis,  they  are  ±  to  MN,  they  are  J.  to  BD.  §  462 

Therefore  AB  and  CLD  are  parallel.  q.  e.  d. 

484.  Cor.  1.  If  one  of  two  parallel  lines  is  perpendicular  to 
a  plane,  the  other  is  also  perpendicular  to  the  plane. 

For,  if  AB  and  CD  are  II,  and  AB  is  ±  to  the  plane  MN, 
and  if  through  any  point  0  of  CD  a  line  is  drawn 
±  to  MN,  it  will  be  II  to  AB  (§  483).  Since 
through  the  point  0  only  one  line  can  be  drawn 
II  to  AB  (§  101),  CD  will  coincide  with  this  J_ 
and  be  J_  to  MN. 

485.  Cor.  2.  If  two  straight  lines  AB  and  EF 
are  parallel  to  a  third  line  CD,  they  are  parallel 
to  each  other.  For,  a  plane  MN  _L  to  CD,  is  J_ 
to  AB  and  ^i^(§484).  Hence  AB  and  EF, 
being  ±  to  MN,  are  parallel  (§  488). 


A 

G 

M 

0 

n 

^=^^ 

^ 

/B 

Df 

N 

parallel  lines  and  planes. 

Parallel  Lines  and  Planes 

Proposition  VIII.     Theorem. 

486.  //  two  straight  lines  are  parallel,  every  plane 
containing  one  of  the  lines,  and  only  one,  is  parallel 
to  the  other  line. 

4 ™™B 


Let  AB  and  CD  be  two  parallel  lines,  and  MN  any 
plane  containing:  CD  and  not  AB. 

To  prove  AB  and  MN  parallel. 

Proof.  The  lines  AB  and  CJD  are  in  the  same  plane  A  BCD, 
which  intersects  the  plane  MNin  the' line  CD. 

Since  AB  is  in  the  plane  AD,  it  must  meet  the  plane  3IN, 
if  at  all,  in  a  point  common  to  the  two  planes ;  that  is,  in 
a  point  of  their  intersection  CD.  But  since  AB  is  II  to  CD, 
it  cannot  meet  CD.  Therefore  AB  cannot  meet  the  plane 
MN,  and  hence  is  II  to  MN.  q.  e.  d. 

487.  Cor.  1.  Through  a  given  straight  line  a  plane  can  be 
passed  parallel  to  any  other  given  straight 
line  in  space.  For,  if  a  plane  is  passed 
through  one  of  the  lines  AB  and  any  point 
G  of  the  other  line  CD,  and  a  line  CE  is 
drawn  in  this  plane  11  to  AB,  the  plane  MN 
determined  by  CD  and  CE  is  II  to  .45.  §486 


254 


SOLID    GEOMETEY. —  BOOK    VI. 


488.    Cor.  2.    Through  a  given  point  a  plane  can  he  passed 
parallel  to  any  two  given  straight  lines  in  space. 

For,  if  0  is  the  given  point,  and  AB  and 
CD  the  given  lines,  by  drawing  through  0  a 
line  A'B'  II  to  AB  in  the  plane  determined 
by  AB  and  0,  and  also  a  line  CD'  II  to  CD 
in  the  plane  determined  by  CD  and  0,  we 
shall  have  two  lines  A'B^  and  C^D^  which 
determine  a  plane  passing  through  0  and 
liaes  AB  and  CD. 


I  trj^Q—p'  I 


to  each  of  the 
§486 


Proposition  IX.     Theorem. 


489.  If  a  given  straight  line  is  parallel  to  a  given 

plane,  the  intersection  of  the  given  plane  with  any 

plane  passed  through  the  given  line  is  parallel  to 

that  line. 

A  B 

\1| 


Let  the  line  AB  be  parallel  to  the  plane  MN,  and  let 
CD  be  the  intersection  of  MN  with  any  plane  passed 
through  AB. 

To  prove  AB  and  CD  parallel. 

Proof.  The  lines  AB  and  CD  are  in  the  same  plane  ABCD, 
and  therefore  if  the  line  AB  meets  the  line  CD,  it  must  meet 
the  plane  MN. 

But  AB  is  by  hypothesis  II  to  MN,  and  therefore  cannot 
meet  it ;  that  is,  it  cannot  meet  CD,  however  far  they  may  be 
produced. 


Hence  AB  and  CD  are  parallel. 


Q.E.  D. 


PARALLEL  LINES  AND  PLANES. 


2d5 


490.  Cor.  If  a  given  straight  line  and  a  plane  are  'parallel, 
a  parallel  to  the  given  line  drawn  through  any  point  of  the 
plane  lies  in  the  plane. 

For  the  plane  determined  by  the  given  line  AB  and  any 
point  (7  of  the  plane  cuts  MN  in  a  line  CD  II  to  AB  (§  489)  ; 
but  through  C  only  one  parallel  to  AB  can  be  drawn  (§  101)  ; 
therefore  a  line  drawn  through  C  II  to  AB  coincides  with  CD, 
and  hence  lies  in  the  plane  MN. 


Proposition  X.     Theorem. 


491.  Two  planes  perpendicular  to  the  same  straight 
line  are  parallel. 


Let  MN  and  PQ  be  two  planes  perpendicular  to  the 
straight  line  AB. 

To  prove  MN  and  PQ  parallel. 

Proof.    ^i*v''and  PQ  cannot  meet.     For  if  they  could  meet, 

we  should  have  two  planes  from  a  point  of  their  intersection 

J^  to  the  same  straight  line.     But  this  is  impossible,         §  475 

{through  a  given  point  vnthout  a  straight  line,  only  one  plane  can  he  passed 

±  to  the  given  line). 

Therefore  JfiVand  PQ  are  parallel. 

Q.  E.  D. 


Ex.  480.  Find  the  locus  of  a  point  in  space  equidistant  from  two 
given  parallel  planes. 

Ex.  481.  Find  the  locus  of  a  point  in  apace  equidistant  from  two 
given  points  and  also  equidistant  from  two  given  parallel  planes. 


256  SOLID    GEOMETEY.  —  BOOK    VI. 


Proposition  XI.     Theorem. 

492.   The  intersections   of  two  parallel  planes  hy  a 

third  plane  are  parallel  lines. 

R 


\y 


Let  the  parallel  planes  MN  and  PQ  be  cut  by  B8. 

To  prove  the  intersections  AB  and  CD  parallel. 

Proof.    AB  and  CD  are  in  the  same  plane  R8. 

They  are  also  in  the  parallel  planes  MN  and  FQ,  which 

cannot  meet  however  far  they  extend. 

Therefore  AB  and  CD  cannot  meet,  and  are  parallel. 

a  E.  D. 

493.  Cor.  1.  Parallel  lines  included  between  parallel  planes 
are  equal. 

For,  if  the  lines  ^(7  and  BD  are  II,  the  plane  of  these  lines 
will  intersect  JfiV^  and  PQ  in  the  II  lines  AB  and  CD  (§  492). 
Hence  ABDCis  a  parallelogram,  and  AC  said  BD  are  equal. 

494.  Cor.  2.  Tivo  parallel  planes  are  everywhere  equally 
distant. 

For'Js  dropped  from  any  points  in  MN  to  PQ  measure  the 
distances  of  these  points  from  PQ.  But  these  Js  are  parallel 
(§  483),  and  hence  equal  (§  493).  Therefore  all  points  in  MN 
are  equidistant  from  PQ. 


PAKALLEL  LINES  AND  PLANES. 


257 


Proposition  XII.     Theorem. 

495.  A  straight  line  perpendicular  to   one  of  two 
parallel  planes  is  perpendicular  to  the  other. 
M 


^^'""^^  ]D..[.r-~-~ ' 

f-«::;3     / 

p 

\       N 

/^    ^ 

..—-'-"""2 

i     1 

Let  AB  be  perpendicular  and  PQ  parallel  to  MN. 

To  prove  that  AB  is  perpendicular  to  FQ. 

Proof.  Pass  through  the  line  AB  any  two  planes  intersect- 
ing JO^in  the  lines  ^Cand  AD,  and  PQ  in  BE  and  BF. 
Then  .4(7  and  AD  are  II  to  BE  and  .5i^  respectively.      §  492 

But  AB  is  ±  to  4Cand  4Z>  (§  462),  and  is  therefore  ±  to 
their  parallels  BE  and  BF.  §  102 

Therefore,  AB  is  ±  to  PQ.  §  472 

Q.E.D. 

496.  Cor.  1.  Through  a  given  point  A  one  plane,  and  only 
07ie,  can  be  drawn  parallel  to  a  given  plane  FQ.  For,  if  a  line 
is  drawn  from  4  X  to  FQ,  a  plane  passing  through  A  Xto 
this  line  is  II  to  FQ  (§  491) ;  and  since  only  one  plane  can  be 
drawn  through  a  point  X  to  a  given  line  (§  474),  only  one 
plane  can  be  drawn  through  A  II  to  FQ. 

497.  Cor.  2.  Jf  two  intersecting  lines  AC  and  AD  are  each 
parallel  to  a  plane  FQ,  the  plane  of  these  lines  MN  is  parallel 
to  FQ.  For  draw  AB  Xio  FQ,  and  through  the  point  B 
draw  BE  and  BF  W  to  AC  and  AD.  Then  BE  and  BF  lie 
in  the  plane  FQ  (§  490).  Hence  AB  is  ±  to  BE  and  BF. 
Therefore  AB  is  ±  to  4C  and  AD  (?  102),  and  hence  to  the 
plane  MN  {^  472).    Hence  JfiVand  FQ  are  parallel.      §  491 


258 


SOLID    GEOMETRY. BOOK    VI. 


Proposition  XIII.     Theorem. 

498.  If  two  angles  not  in  the  same  plane  have  their 
sides  respectively  parallel  and  lying  in  the  same  di- 
rection, they  are  equal,  and  their  planes  are  parallel. 


Let  the  angles  A  and  A'  be  respectively  in  the  planes 
MN and  PQ  and  have  AD  parallel  to  AD'  and  AC  par- 
allel to  A'C    and  lying  in  the  same  direction. 

To  prove        /.  A  =  Z.A\  and  MN  II  to  PQ. 

Proof.  Take  AB  and  A^ D^  equal,  also  J.Cand  ^'(7' equal. 

Join  AA\  DD\  CC\  CD,  C'D\ 

Since  AD  is  equal  and  II  to  A^D\  the  figure  ADD^A^  is  a 
parallelogram,  and  AA^  =  DD\  §  182 

In  like  manner  AA^  =  CC\ 

Also,  since  (7(7'  and  DD^  are  each  II  to  AA\  and  equal  to 
AA\  they  are  II  and  equal. 

Therefore  CDD'O'  is  a  parallelogram,  and  CD  =  CD'. 

.-.  A  ADC=---  A  A'D'C,  and  Z  A--=ZA\         §  160 

Also,  since  FQ  is  II  to  each  of  the  lines  AC  and  AD  (§  486). 
FQ  is  II  to  the  plane  of  these  lines  MN{^  497).  q.  e.  d. 


PARALLEL  LINES  AND  PLANES. 


259 


Proposition  XIV.     Theorem. 

499.  If  two  straight  lines  are  intersected  by  three 
parallel  planes,  their  corresponding  segments  are 
proportional, 

M C 

^' --- ^^/^ 

Let  AB  and  CD  be  intersected  by  the  parallel  planes 
MN,  PQy  RS,  in  the  points  A,  E,  B,  and  C,  F,  D. 

To  prove  AE :  EB  =  CF:  FD. 

Proof.        Draw  AD  cutting  the  plane  PQ  in  G. 

Join  EO  and  EG. 

Then  EG  is  II  to  BD,  and  GE'\^  II  to  AC.  §  492 

.\AE'.EB  =  AG:  GD,  §309 

and  CF:FD=AG.GD. 

:.AE:EB=CF:FI). 

aE.  D. 

Ex.  482.  The  line  AB  meets  three  parallel  planes  in  the  points  A, 
E,  B  ;  and  the  line  CD  meets  the  same  planes  in  the  points  C,  F,  D.  If 
AE=  6  inches,  BE=S  inches,  (7i>  =  12  inches,  compute  CFand  FD. 

Ex.  483.  To  draw  a  perpendicular  to  a  given  plane  from  a  given 
point  without  it. 

Ex.  484.  To  erect  a  perpendicular  to  a  given  plane  at  a  given  point 
in  it. 


260 


SOLID   GEOMETRY.  —  BOOK   VI. 


Dihedral  Angles. 


500.  The  opening  between  two  intersecting  planes  is  called 
a  dihedral  angle. 

The  line  of  intersection  AB  of  the  planes  is  the  edge,  and 
the  planes  MA  and  NB  are  the /aces  of  the  dihedral  angle. 

501.  A  dihedral  angle  is  designated  by  its  edge,  or  by  its 
two  faces  and  its  edge.     Thus,  the 
dihedral  angle  in  the  margin  may  be 
designated  by  AB,  or  by  M-AB-N. 

502.  In  order  to  have  a  clear 
notion  of  the  magnitude  of  the  di- 
hedral angle  AB,  suppose  a  plane 
at  first  in  coincidence  with  MA  to 
turn  about  the  edge  AB,  in  the 
direction  indicated  by  the  arrow, 
until  it  coincides  with  the  face  NB.  The  amount  of  rotation 
of  this  plane  is  the  dihedral  angle  AB. 

503.  Two  dihedral  angles  are  equal  when  they  can  be  made 
to  coincide.  I 

504.  Two  dihedral  angles  M-AB-JSf  and  JSf-AB-P  are  adja- 
cent if  they  have  a  common  edge     ^  ^ 
AB,  and  a  common  face  NAB,  be- 
tween them. 

505.  When  a  plane  meets  another 
plane  and  makes  the  adjacent  dihe- 
dral angles  equal,  each  of  these  an- 
gles is  called  a  right  dihedral  angle. 

506.  A  plane  is  perpendicular  to  another  plane  if  it  forms 
with  this  second  plane  a  right  dihedral  angle. 


DIHEDRAL   ANGLES. 

507.  Two  vei^tical  dihedral  angles  are  angles  that  have  the 
same  edge  and  the  faces  of  the  one  are  the  prolongations  of 
the  faces  of  the  other. 

508.  Dihedral  angles  are  acute,  obtuse^  complementary, 
supplementary,  under  the  same  conditions  as  plane  angles 
are  acute,  obtuse,  complementary,  supplementary,  respec- 
tively. 

509.  The  demonstrations  of  many  properties  of  dihedral 
angles  are  identically  the  same  as  the  demonstrations  of  anal- 
ogous theorems  of  plane  angles. 

The  following  are  examples  : 

1.  If  a  plane  meets  another  plane,  it  forms,  with  it  two 
adjacent  dihedral  angles  whose  sum  is  equal  to  two  right 
dihedral  angles. 

2.  If  the  sum  of  two  adjacent  dihedral  angles  is  equal  to 
two  right  dihedral  angles,  their  exterior  faces  are  in  the  same 
plane. 

3.  If  two  planes  intersect  each  other,  their  vertical  dihedral 
angles  are  equal. 

4.  If  a  plane  intersects  two  parallel  planes,  the  exterior- 
interior  dihedral  angles  are  equal ;  the  alternate-interior  dihe- 
dral angles  are  equal ;  the  two  interior  dihedral  angles  on  the 
same  side  of  the  secant  plane  are  supplements  of  each  other. 

5.  When  two  planes  are  cut  by  a  third  plane,  if  the  exterior- 
interior  dihedral  angles  are  equal,  or  the  alternate-interior 
dihedral  angles  are  equal,  and  the  edges  of  the  dihedral  angles 
thus  formed  are  parallel,  the  two  planes  are  parallel. 

6.  Two  dihedral  angles  whose  faces  are  parallel  each  to 
each  are  either  equal  or  supplementary. 

7.  Two  dihedral  angles  whose  faces  are  perpendicular  each 
to  each  are  either  equal  or  supplementary. 


262 


SOLID    GEOMETRY. — BOOK    VI. 


Measure  of  Dihedral  Angles. 

510.  The  plane  angle  of  a  dihedral  angle  is  the  plane  angle 
formed  by  two  straight  lines,  one  in  each  plane,  perpendicular 
to  the  edge  at  the  same  point. 

511.  The  plane  angle  of  a  dihedral  angle  has  the  same 
magnitude  from  whatever  point  in  the  edge  the 
perpendiculars  are  drawn.  For  any  two  such 
angles,  as  CAB,  GIIT,  have  their  sides  respec- 
tively parallel  (§  100),  and  hence  are  equal 
(§498). 

Proposition  XV.     Theorem. 

512.  Two  dihedral  angles  are  equal  if  their  plane 
angles  are  equal. 


\ 

c 

H 

F 

G 

E 

Let  the  two  plane  angles  ABD  and  A'B'D'  of  the  two 
dihedral  angles  CB  and  C'B'  be  equal. 

To  prove  the  dihedral  angles  CB  and  C'B^  equal. 

Proof.  Apply  B'C^  to  BC,  making  the  plane  angle  A'B'B' 
coincide  with  its  equal  ABD. 

The  line  B^C  being  ±  to  the  plane  A'B'B'  will  likewise  be 
X  to  the  plane  ABB  at  B,  and  take  the  direction  BC,  since 
at  B  only  one  _L  can  be  erected  to  this  plane.  §  476 

The  two  planes  A'B'C  and  ABC,  having  in  common  two 
intersecting  lines  AB  and  BC,  coincide.  §  460 

In  like  manner  the  planes  B'B'C  and  Z>^C  coincide. 

Therefore  the  two  dihedral  angles  coincide  and  are  equal. 


Q.  E.  D. 


DIHEDRAL   ANGLES. 


263 


Proposition  XVI.    Theorem. 

* 

513.  Two  dihedral  angles  have  tJie  same  ratio  as 

their  plane  angles. 

B' 
A  "      " 


Case  I.    When  the  plane  angles  are  commensurahle. 

Let  A-BC-D  and  A'-B'a-D'be  two  dihedral  angles,  and 
let  their  plane  angles  ABD  and  A'B'D'  be  commensur- 
able. 

To  prove  A-BC-D  :  A'-B'C'-D^  =  /.  ABD  :  Z  A'B'D'. 

Proof.  Suppose  the  A  ABD  and  A'B'D'  have  a  common 
measure,  which  is  contained  three  times  in  Z.  ABD  and  five 
times  in  Z  A'B'D'. 

Then  "      A  ABD'.AA'B'D'^2>:b. 

Apply  this  measure  to  Z  ABD  and  Z  A'B'D',  and  through 
the  lines  of  division  and  the  edges  ^Cand  B'C  pass  planes. 

These  planes  divide  A-BC-D  into  three  parts,  and  A'-B'C'D' 
into  five  parts,  all  equal  because  they  have  equal  plane  angles. 

Therefore       A-BC-D  :  A'-B'C'-D'  =  3:5. 

Therefore  A-BC-D  :  A'-B'C'-D'  -  Z  ABD  :  Z  A'B'D'. 


264 


SOLID    GEOMETRY.  —  BOOK    VI. 


Case  II.    When  the  plane  angles  are  incommevisurahle. 


Let  A-BC-D,  A'-B'C-D'  be  dihedral  angles,  and  let  their 
plane  angles  ABD,  A'B'D'  be  incommensurable. 

To  prove  A-BC-D  :  A'-B^C'-D'  ^ZaBD.Z  A'B'D\ 

Proof.  Divide  the  Z  ABD  into  any  number  of  equal  parts, 
and  apply  one  of  these  parts  to  the  Z.  A'B'D^  as  a  measure. 

Since  ABD  and  A'B'D'  are  incommensurable,  a  certain 
number  of  these  parts  will  form  the  Z  A'B'U,  leaving  a 
remainder  JSB'D',  less  than  one  of  the  parts. 

Pass  a  plane  through  ^'^and  B'C 

Since  the  plane  angles  of  the  dihedral  Singles. A-BC-D  and 
A'-B'C'-BJ  Sive  commensurable,  we  have  by  Case  I., 

A-BC-D  :  A'-B'C'-U=Z  ABD  :  Z  A'B'K 

If  the  unit  of  measure  is  indefinitely  diminished,  these  ratios 
continue  equal, and  approach  indefinitely  the  limiting  ratios, 

A-BC-D  :  A'-B'C'-D,  and  Z  ABD  :  Z  A'B'D'. 

.-.  A-BC-D  :  A'-B'C'-D'  -  Z  ABD  :  Z  A'B'D'.    §  260 

Q.  E.  D. 

514.  Scholium.  The  plane  angle  is  taken  as  the  measure 
of  the  dihedral  angle.     (Compare  §  262.) 


DIHEDRAL   ANGLES. 


265 


Planes  Perpendiculab,  to  Each  Otheb. 
Proposition  XVII.     Theorem. 

515.  If  two  'planes  are  perpendicular  to  each  other, 
a  straight  line  drawn  in  one  of  them  perpendicular 
to  their  intersection  is  perpendicular  to  the  other. 


A  N 

Let  the  plane  PAB  be  perpendicular  ta  MN,  and  let 
CD  be  drawn  in  PAB  perpendicular  to  their  intersec- 
tion AB. 

To  prove  CD  perpendicular  to  MN. 

Proof.   In  the  plane  MN Am\N  DEI.  to  AB  at  D. 

Then  CDE  is  the  plane  angle  of  the  right  dihedral  angle 
P-AB-N,  and  is  therefore  a  right  angle. 

By  construction  CD  A  is  a  right  angle. 

Therefore  CD  is  X  to  DA  and  DE  at  their  point  of  inter- 
section, and  consequently  -L  to  their  plane  MN.  §  472 

Q.  E.  D. 

516.  Cor.  1.  If  two  planes  are  perpendicular  to  each  other ^ 
a  perpendicular  to  one  of  them  at  any  point  of  their  intersection 
will  lie  in  the  other. 

For,  a  line  CD  drawn  in  the  plane  PAB  _L  to  AB  at  the 
point  D  will  be  X  to  MN  {%  515).  But  at  the  point  D  only 
one  J_  can  be  drawn  to  MN  (§  476).  Therefore  a  X  to  MN 
erected  at  D  will  coincide  with  CD  and  lie  in  the  plane  PAB, 


266 


SOLID    GEOMETRY 


BOOK    VI. 


617.  Cor.  2.  If  two  planes  are  perpendicular  to  each  other, 
a  perpendicular  to  one  of  them  from  any  point  of  the  other  will 
lie  in  the  other. 

For,  a  line  CD  drawn  in  the  plane  PAB  from  the  point  Q 
L  to  AB  will  be  ±  to  MN{^  515).  But  from  the  point  C  only 
one  ±  can  be  drawn  to  MN  (§  476).  Therefore  a  ±  to  MN 
drawn  from  Cwill  coincide  with  CD  and  lie  in  PAB. 


Proposition  XVIII.     Theorem. 

518.  If  a  straight  line  is  perpendicular  to  a  plane, 
every  plane  passed  through  the  line  is  perpendicu- 
lar to  the  first  plane. 

cA 


li 


aX 


r-S        I 


Let  CD  be  perpendicular  to  MN,   and  PAB   be  any 
plane  passed  through  CD  intersecting  MN  in  AB. 

To  prove  the  plane  PAB  perpendicular  to  the  plane  MN. 

Proof.    Draw  DE  in  the  plane  MN,  and  J.  to  AB. 

Since  CD  is  ±  to  MN,  it  is  JL  to  AB. 

Therefore  Z  CDE  is  the  plane  angle  of  P-AB-N 

But  Z  CDEh  a  right  angle, 

and  therefore  PAB  is  ±  to  MN  §  514 

Q.  E.  D. 

.  519.    Cor.  A  plane  perpendicular  to  the  edge  of  a  dihedral 
angle  is  perpendicular  to  each  of  its  faces. 


DIHEDRAL   ANGLES. 


267 


Proposition  XIX.     Theorem. 

520.  If  two  intersecting  -planes  are  each  perpendicu- 
lar to  a  third  plane,  their  intersection  is  also  perpen- 
dicular to  that  plane. 


Let  the  planes  BD  and  BC  intersecting  in  the  line 
AB  be  perpendicular  to  the  plane  PQ. 

To  prove  AB  perpendicular  to  the  plane  PQ. 

Proof.   A  X  erected  to  FQ  at  B,  a  point  common  to  the 
three  planes,  will  lie  in  the  two  planes  ^Cand  £D.       §  516 

And  since  this  JL  lies  in  both  the  planes  BC  and  BI),  it 
must  coincide  with  their  intersection  AB. 


.-.  AB  is  ±  to  the  plane  PQ. 


aE.  D. 


521.  Cor.  1.  If  a  plane  PQ  is  perpendicular  to  each  of  two 
intersecting!  planes  ABC  and  ABI),  it  is  perpendicular  to  their 
intersection  AB. 

522.  Cor.  2.  If  a  plane  PQ  is  perpendicular  to  two  planes 
ABC  and  ABD,  which  include  a  right  dihedral  angle,  the 
intersection  of  any  two  of  these  planes  is  perpendicular  to  the 
third  plane,  and  each  of  the  three  intersections  is  popendicular 
to  the  other  two. 


268  SOLID    GEOMETRY.  —  BOOK    VI. 

Proposition  XX.     Theorem. 

523.  Through  a  given  straight  line  not  perpendicu- 
lar to  a  plane,  one  plane,  and  only  one,  can  be  passed 
perpendicular  to  the  given  plane. 

J? 


Let  AB  be  the  given  line  not  perpendicular  to  the 
plane  MN. 

To  prove  that  one  plane  can  he  parsed  through  AB  perpen- 
dicular to  MN,  and  only  one. 

Proof.  From  any  point  A  of  AB  draw  AC l^io  MN, 

and  through  AB  and  J^Cpass  a  plane  AD. 

The  plane  AD  is  _L  to  MN,  since  it  passes  through  AC,  a. 
line  ±  to  MN  §  518 

Moreover,  if  two  planes  could  be  passed  through  AB  J_  t(y 
the  plane  MN,  their  intersection  AB  would  be  ±  to  MN.  §  520 

But  this  is  impossible,  since  AB  is  by  hypothesis  oblique  to 
the  plane  MN. 

Hence  through  AB  only  one  plane  can  be  passed  X  to  3fN. 

OLE  D. 

624.  Cor.  If  a  straight  line  is  oblique  to  a  plane,  its  projec- 
tion IS  a  st7^aight  line. 

For,  the  plane  passed  through  ^^  _L  to  MN  contains  all 
the  Js  let  fall  from  different  points  of  AB  upon  MN  (§  516). 
Therefore  the  intersection  CD  of  these  planes  is  the  locus  of  the 
projections  of  the  points  in  AB.  But  the  intersection  CD\s  a 
straight  line  ;  that  is,  the  projection  of  AB  is  a,  straight  line. 


DIHEDRAL    ANGLES.  269 


Proposition  XXI.     Theorem. 

625.  Every  point  in  a  plane  which  bisects  a  dihe- 
dral angle  is  equidistant  from  the  faces  of  the 
angle. 


A 
Let  plane  AM  bisect  the  dihedral  angle  formed  by 
the  planes  AD  and  AC ;  and  let  PE  and  FF  be  perpen- 
diculars drawn  from  any  point  P  in  the  plane  AM  to 
the  planes  AO  and  AD. 

To  prove  FE=PF. 

Proof.    Through  FE  and  FF  pass  a  plane  intersecting  the 
planes  ^(7 and  AD  in  the  lines  O^and  OF,  and  join  FO. 

The  plane  FEE  is  ±  to  ^  (7  and  to  AD.  §  518 

Hence  the  plane  FEE  is  JL  to  their  intersection  AO.  ^  521 
.:/.FOE=AFOE, 

[being  measures  respectively  of  the  equal  dihedral  A  M-OA-  C  and  M- OA-D). 

.'.  rt.  A  FOE=  rt.  A  FOE.  §  148 

.'.FE=FF. 

a  E.  D. 


vEx.  485.   Find  the  locus  of  a  point  in  space  equidistant  from  three 
given  points  not  in  a  straight  line. 

Ex.  486.  Given  two  points  A  and  B  on  the  same  side  of  a  given  plane 
MN;  find  a  point  in  this  plane  such  that  the  sum  of  its  distances  from 
A  and  B  shall  be  a  minimum. 


270  SOLID    GEOMETRY.  —  BOOK    VI. 

Angle  of  a  Straight  Line  and  a  Plane. 

Proposition  XXIL     Theorem. 

526,  The  acute  angle  which  a  straight  line  makes 
with  its  own  projection  upon  a  plane  is  the  least  an- 
gle which  it  makes  with  any  line  of  the  plane. 


Let  BA  meet  the  plane  MN  at  A,  and  let  AC  "be  its 
projection  upon  the  plane  MN,  and  AD  any  other  line 
drawn  through  A  in  the  plane. 

To  prove  A  BA  C  less  than  Z  BAB. 

Proof.  Take  AD  =  AC,  and  join  BB. 

In  the  A  B AC SiTid  BAB 

BA  =  BA,  AC=  AB,  but  BC<  BB.  §  477 

.-.  Z  BACis  less  than  Z  BAB,  §  153 

Q.  E.  a 

527.  Scholium.  If  the  straight  line  AC  turns  about  the 
point  A,  the  angle  ^^C  increases ;  it  is  a  right  angle  when 
^C  is  perpendicular  to  its  initial  position;  then  it  becomes 
obtuse,  and  reaches  its  maximum  value  when  AC  falls  upon 
AGHhe  prolongation  of  CA.  Afterwards  the  angle  passes 
through  the  same  values  in  reverse  order. 


DIHEDRAL   ANGLES.  271 

* 

A  Perpendicular  between  Two  Straight  Lines. 
Proposition  XXIII.     Theorem. 


^^^^  Between   tivo  straight   lines  not  in  tl%e  same 

plane,  one  common  perpendicular  can  be  drawn,  and 

only  I  one, 

C  E  D 


M_Al_,/ ! ...; ^ 

Let  AB  and  CD  be  the  given  lines. 

To  prove  that  one  common  perpendicular  can  be  drawn  be- 
tweert  them^  and  only  one. 

Proof.  Through  any  point  B  of  AB  draw  BG  II  to  CD,  and 
let  JfiVbe  the  plane  determined  by  AB  and  BG. 

Through  CD  pass  the  plane  CD'  X  to  MN,  and  intersecting 
AB  at  C". 

At  C"  erect  a  ±  C'Cto  the  plane  MN.  C'C  will  lie  in  the 
plane  CD  (§  516),  and  be  X  to  AB  and  CD'  (§  462). 

Since  CCis  ±  to  CD',  it  is  JL  to  CD  (§  102). 

Hence  CC  is  a  common  perpendicular  to  CD  and  AB. 

Moreover,  CC  is  the  only  common  perpendicular. 

For,  if  any  other  line  UB  could  be'±  to  CD  and  AB,  it 
would  be  ±  to  BG  and  AB  (§  102),  and  hence  J.  to  MK 

But  .EB:  in  the  plane  CD'  and  J.  to  CD',  is  JL  to  MN 

(§  515),  and  we  should  have  two  Js  from  U  to  MN. 

But  this  is  impossible.  §  476 

Hence  CC  is  the  only  common  ±  to  CD  and  AB. 

an.  D. 


272  SOLID    GEOMETRY,  —  BOOK   VI. 

Polyhedral  Angles. 

529.  A  polyhedral  angle  is  the  opening  of  three  or  more 
planes  which  meet  at  a  common  point. 

530.  The  common  point  8  is  the  vertex  of  the  angle,  and  the 
intersections  of  the  planes  8A,  SB,  etc.,  are  « 

its  edges;  the  portions  of  the  planes  included  A 

between  the  edges  are  its  faces,  and  the  angles  /7  \ 

formed  by  the  edges  are  it^face  angles.  a  /-.       \ 

'/-  --------Ac 

531.  The  magnitude  of  a  polyhedral  angle         B  \ 
depends  upon  the  relative  position  of  its  faces,  and  not  upon 
their  extent. 

532.  In  a  polyhedral  angle,  every  two  adjacent  edges  form 
a  face  angle,  and  every  two  adjacent  faces  form  a  dihedral 
angle.  These  face  angles  and  dihedral  angles  are  the  parts 
of  the  polyhedral  angle. 

533.  Two  polyhedral  angles  ?                ^ 
can  be    made   to    coincide  and  yy  \              /7  \ 
are  equal  if  their  corresponding  A//      \    X//      \ 
parts  are   equal  and    arranged  /  l/'^   ''"-■vAg    A  "■"---Ap' 
in  the  same  order.  ^B           ^      B'          A 

534.  A  polyhedral  angle  is  convex  if  any  section  made  by 
a  plane  cutting  all  its  faces  is  a  convex  polygon. 

535.  A  polyhedral  angle  is  called  trihedral,  tetrahedral,  etc., 
according  as  it  has  three  isices,  four  faces,  etc. 

536.  A  trihedral  angle  is  called  rectangular,  bi-rectangular, 
tri-rectangular,  according  as  it  has  one,  two,  or  three  right 
dihedral  angles. 

Two  adjacent  walls  and  the  floor  of  a  rectangular  room  form 
a  tri-rectangular  trihedral  angle. 

537.  A  trihedral  angle  is  called  isosceles  if  it  has  two  of  its 
face  angles  equal. 


SYMMETRICAL    POLYHEDRAL   ANGLES. 


273 


Symmetrical  Polyhedral  Angles. 

538.  If  the  edges  of  a  given  polyhedral  angle  8-ABCD 
are  produced  through  the  vertex  S,  another  polyhedral  angle 
S-A'B'C'D'  is  formed,  symmetrical  with  respect  to  S-AB CD. 
The  face  angles  A8B, 
B80,  etc.,  are  equal 
respectively  to  the  face 
angles  A'8B\  B'8C', 
etc.,  since  they  are  ver- 
tical angles. 

Also  the  dihedral 
angles  whose  edges  are 
8A,  8B,  etc.,  are  equal  ^ 

respectively  to  the  dihedral  angles  whose  edges  are  8A',  8B\ 
etc.,  since  they  are  vertical  dihedral  angles.  (The  second 
figure  shows  a  pair  of  vertical  dihedral  angles.) 

The  edges  of  8-ABCD  are  arranged  from  left  to  right  in  the 
order  8B,  8C,  8D,  but  the  edges  of  8-A'B'C'D'  are  arranged 
from  right  to  left  in  the  order  8B\  8C',  8D' ;  that  is,  in  an 
order  the  reverse  of  the  order  of  the  edges  in  8-ABCD. 

Two  symmetrical  polyhedral  angles,  therefore,  have  all  their 
parts  equal,  each  to  each,  but  arranged  in  reverse  order. 

In  general,  two  symmetrical  polyhedral  angles  are  not  super- 
posable.  Thus,  if  the  trihedral  angle  8-A'B'C'  is  made  to 
turn  180°  about  the  bisector  xy  of  the  angle  A'8C,  the  side 
8A'  will  coincide  with  8C,  8G'  with  8A,  and 
the  face  A'8C'  with  A8C;  but  the  dihedral 
angle  8 A,  and  hence  the  dihedral  angle  8A', 
not  being  equal  to  8C,  the  plane  A'8B'  will 
not  coincide  with  B8C;  and,  for  a  similar 
reason,  the  plane  C'8B^  will  not  coincide  with 
A8B.  Hence  the  edge  8B'  takes  some  posi- 
tion 8B"  not  coincident  with  8B ;  that  is,  the  trihedral  angles 
are  not  superposable. 


J 


074. 


SOLID    GEOMETRY.  —  BOOK    VI. 


Proposition  XXIV.     Theorem. 


X      539.  The  sum  of  any  two  face  angles  of  a  trihedral 
angle  is  greater  than  the  third  face  angle. 

S 


In  the  trihedral  angle  S-ABG  let  the   angle  ASC  be 
greater  than  A8B  or  BSG. 

To  prove  Z.  A8B  -f-  Z  BBC  greater  than  Z  ASC. 

Proof.    In  A8C  draw  SD,  making  Z  ASD  =  Z  A8B. 

Through  any  point  D  of  BB  draw  ABCixi  the  plane  ABC 

Take  SB  =  BD. 

Pass  a  plane  through  the  line  ^Cand  the  point  B. 

In  the  A  ABB  and  ABB, 

AB=  AB,  BB  =  BB,  and  Z  ABB  =  Z  ABB. 

.-.  A  ABB  =  A  ABB  (§  150),  and  AB  =  AB. 
InihQAABC;       AB-\-BC>  AC  §137 

But  AB  =AB 

By  subtraction,  BC>  BC 

In  the  A  BBC  and  BBC, 

BC=  BC,  and  BB  =  BB,  but  BC>  BC. 
Therefore  Z  BBC  is  greater  than  Z  BBC.         §  153 
:•.  A  ABB  +  BBC  are  greater  than  A  ABB  +  BBC 
That  is,  A  ABB  +  BBC  are  greater  than  Z  ^/^C.  . 

f  Q.  E.  D. 


POLYHEDKAL   ANGLES.  275 


Proposition  XXV.    Theorem. 

540.  The  sum  of  the  face  angles  of  any  convex  poly- 
hedral angle  is  less  than  four  right  angles. 


Let  S  be  a  convex  polyhedral  angle,  and  let  its 
faces  be  cut  by  a  plane,  making  the  section  ABCBE 
a  convex  polygon. 

To  prove  Z  A8B  -\-  Z  B8C,  etc.,  less  than  four  rt.  A. 

Proof.  From  any  point  0  within  tbe  polygon  draw  OA,  OB^ 
OC,  OD,  OE. 

The  number  of  the  A  having  their  common  vertex  at  0  will 
be  tl|3  same  as  the  number  having  their  common  vertex  at  8. 

Therefore  the  sum  of  all  the  A  of  the  A  having  the  common 
vertex  at  8  is  equal  to  the  sum  of  all  the  A  of  the  A  having 
the  common  vertex  at  0. 

But  in  the  trihedral  A  formed  at  ^,  B,  C,  etc. 

Z  8AE+  Z  8AB  is  greater  than  Z  EAB, 
and  Z  8BA  +  Z  8BC  is  greater  than  Z  ABC,  etc.  §  539 

Hence  the  sum  of  the  A  at  the  bases  of  the  A  whose  com- 
mon vertex  is  8  is  greater  than  the  sum  of  the  A  at  the  bases 
of  the  A  whose  common  vertex  is  0. 

Therefore  the  sum  of  the  A  at  the  vertex  8  is  less  than  the 
sum  of  the  A  at  the  vertex  O. 

But  the  sum  of  the  ^  at  0  =  4  rt.  A.  §  92 

Therefore  the  sum  of  the  zi  at  xS'  is  less  than  4  rt.  A. 

Q.E.  D. 


276  SOLID   GEOMETRY.  —  BOOK    VI. 


Proposition  XXVI.     Theorem. 

541.  Two  trihedral  angles  are  equal  or  symmetrical 
when  the  three  face  angles  of  the  one  are  respectively 
equal  to  the  three  face  angles  of  the  other. 


B'  E'  ABE  A  A  E'  B 


In  the  trihedral  angles  S  and  S^  let  the  angles  ASB 
A8C,  BSC,  be  equal  to  the  angles  A'S'B',  A'S'C,  B'S'C, 
respectively. 

To  prove  S-ABQ  and  S^-A^B'C^  equal  or  si/mmetrical. 

Proof.  On  the  edges  of  these  angles  take  the  six  equal  dis- 
tances SA,  SB,  SO,  S'A\  S'B\  S'C\ 

Draw  AB,  BO,  AC,  AB\  BW\  AOK 

The  isosceles  A  BAB,  8 AC,  SBC,  are  equal  respectively 
to  S'A'B',S'A'C',  S'B'C^.  §  150 

AB,  AC,  BCsiYo  equal  respectively  to  A'B',  AC,  B'C. 

.\AABC=AA'B'C'.  §  160 

At  any  point  D  in  SA  draw  BU  and  i)i^  in  the  faces  ASB 
and  ^/S'C  respectively,  and  JL  to  SA. 

These  lines  meet  AB  and  J^C  respectively, 

{since  the  A  SAB  and  8 AC  are  acute,  each  being  one  of  the  equal  A  of  an 

isosceles  A). 

Join  UF. 
OuS'A'  tBke  A'B'  =  AB. 


poly;eedeal  angles.  277 

Draw  Z)'^'  and  D'i^'  in  the  faces  of  A'S'B'  and  A'jS'C 
respectively,  ±  to  S'A',  and  join  JE'F'. 
In  the  rt.  A  ADU  and  A'D'U', 

AD  =  A'B',  and  Z  ZlAU=  Z  B'A'U'. 

.-.  rt.  A  ^i>^=  rt.  A  A'D'E'.  §  149 

.-.  AE=  A'E'  and  BE=  D'E'. 
In  like  manner  we  may  prove  AF=A'F'  and  DF=D'F'. 
Hence  in  the  A  ^^i^and  A'F'F' 
AE=  A'F',  AF=  A'F',  and  Z  EAF=  Z  E'A'F'. 

:.  A  AEF=  A  A'E'F\  and  EF=  E'F\        §  150 
Hence,  in  the  A  EDF  a,nd  E'D'F'  we  have 

ED  =  E'D',  DF=  D'F',  and  EF=  E'F'. 
.'.  A  EDF=  A  ^'X>'i^'  and  Z  EDF=  Z  E'D'F'.  §  160 
Therefore  the  dihedral  angle  B-AS-C  equ&h  dihedral  angle 
D'-A'S'-C, 
{since  A  EDF  and  E'D'F\  the  measures  of  these  dihedral  A,  are  equal). 

In  like  manner  it  may  be  proved  that  the  dihedral  angles 
A- BS-C  and  A-C8-B  are  equal  respectively  to  the  dihedral 
angles  A'-D'S'-C  and  A'~C'8'-B'.  o.  e. d. 

This  demonstration  applies  to  either  of  the  two  figures  de- 
noted by  S'-A'B'C,  which  are  symmetrical  with  respect  to 
each  other.  If  the  first  of  these  figures  is  taken,  /S  and  S'  are 
equal.     If  the  second  is  taken,  S  and  S'  are  symmetrical. 

542.  Scholium.  If  two  trihedral  angles  have  three  face 
angles  of  the  one  equal  to  three  face  angles  of  the  other,  then 
the  dihedral  angles  of  the  one  are  respectively  equal  to  the  dihe- 
dral angles  of  the  other. 

'^x.  487.  An  isosceles  trihedral  angle  and  its  symmetrical  trihedral 
angle  are  superposable. 

^  Ex.  488.  Find  the  locus  of  a  point  equidistant  firom  the  three  edges 
of  a  trihedral  angle. 

*Ex.  489.  Find  the  locus  of  a  p®int  equidistant  from  the  three  faces 
of  a  trihedral  angle. 


BOOK   VII. 
POLYHEDRONS.   CYLINDERS.  AND   CONES. 


Polyhedrons. 

543.  A  polyhedron  is  a  solid  bounded  by  planes. 

The  bounding  planes,  limited  by  each  other,  are  the  faces, 
their  intersections  are  the  edges,  and  the  intersections  of  the 
edges  are  the  vertices,  of  the  polyhedron. 

544.  A  diagonal  of  a  polyhedron  is  a  straight  line  joining 
any  two  vertices  not  in  the  same  face. 

545.  A  polyhedron  of  four  faces  is  called  a  tetrahedron ;  one 
of  six  faces,  a  hexahedrbn;  one  of  eight  faces,  an  octahedron; 
one  of  twelve  faces,  a  dodecahedron;  one  of  twenty  faces,  an 
icosahedron. 

546.  A  polyhedron  is  convex  if  the  section  made  by  any 
plane  cutting  it  is  a  convex  polygon. 

Only  convex  polyhedrons  are  considered  in  this  work. 

547.  The  volume  of  a  solid  is  its  numerical  measure,  referred 
to  another  solid  taken  as  the  unit  of  volume. 

548.  A  polyhedron  of  six  faces,  each  face  a  square,  is  called 
a  cube;  and  the  cube  whose  edge  is  the  linear  unit  is  generally 
taken  as  the  unit  of  volume. 

549.  Two  solids  are  equivalent  if  their  volumes  are  equal. 

550.  Two  polygons  are  parallel  if  their  sides  are  respectively 
parallel. 


POLYHEDRONS. 


279 


Prisms  and  Parallelopipeds. 

551.  A  prism  is  a  polyhedron  of  which  two  opposite  faces, 
called  bases,  are  parallel  polygons,  and  the  other 
faces,  called  lateral  faces,  intersect  in  parallel 
lines,  called  lateral  edges.  The  lateral  edges  are 
equal  (§  493), the  lateral  faces  are  parallelograms 
(168),  and  the  bases  are  equal  (§§  179,  498). 

552.  The  sum  of  the  areas  of  the  lateral  faces  of  a  prism  is 
called  its  lateral  area. 

553.  The  altitude  of  a  prism  is  the  length  of  the  perpen- 
dicular between  the  planes  of  its  bases. 

554.  Prisms  are  called  triangular,  quadrangular,  etc.,  ac- 
cording as  their  bases  are  triangles,  quadrilaterals,  etc. 

555.  A  right  section  of  a  prism  is  a  section  made  by  a  plane 
perpendicular  to  its  lateral  edges. 


Truncated  Prism.  Right  Prism.  Rectangular  Parallelopiped.         Parallelopiped. 

556.  A  truncated  prism  is  the  part  of  a  prism  included 
between  the  base  and  a  section  made  by  a  plane  inclined  to 
the  base  and  cutting  all  the  lateral  edges. 

557.  An  oblique  prism  is  a  prism  whose  lateral  edges  are  not 
perpendicular  to  its  bases ;  a  right  prism  is  a  prism  whose  lat- 
eral edges  are  perpendicular  to  its  bases  ;  a  regular  prism  is  a 
right  prism  whose  bases  are  regular  polygons. 

558.  A  prism  whose  bases  are  parallelograms  is  called  a 
parallelopiped.  If  its  lateral  edges  are  perpendicular  to  the 
bases,  it  is  called  a  right  parallelopiped.  If  its  six  faces  are 
all  rectangles,  it  is  called  a  rectangular  parallelopiped. 


280  SOLID    GEOMETRY.  —  BOOK    VII. 


Proposition  I.     Theorem. 

559.  The  sections  of  a  prism  made  hy  parallel  planes 
are  equal  polygons, 

^"^         J) 


Let  the  prism  AD  be  intersected  by  the  parallel 
planes  GK,  G'K'. 

To  prove  GHIKL  =  G'TT'I'X'L'. 

Proof.  Since  the  intersections  of  two  parallel  planes  by  a 
third  plane  are  parallel  (§  492),  the  sides  GIT,  HI,  IK,  etc., 
are  parallel  respectively  to  the  sides  G^H^,  II'I\  I'K',  etc. 

The  sides  GS,  SI,  IK,  etc.,  are  equal  respectively  to  G^H\ 
Wr,  VK,  etc., 

since  parallel  lines  comprehended  between  parallel  lines  are 
equal.  .  §  180 

The  A  GUI,  HIK,  etc.,  are  equal  respectively  to  A  &WI\ 
WVK,  etc., 

since  two  A  not  in  the  same   plane,  having   their   bides 

respectively  parallel  and   lying   in   the  same  direction,  are 

equal.  §  498 

Therefore  GHIKL  =  G'HI'K'H,  §  203 

because  they  are  mutually  equiangular  and  equilateral. 

Q.  E.  D. 

560.  Cor.  Aut/  section  of  a  prism  parallel  to  the  base  is  equal 
to  the  base;  and  all  right  sections  of  a  prism  are  equal. 


PRISMS.  281 


Proposition  II.     Theorem. 

561.  The  lateral  area  of  a  prism  is  equal  to  the 
■product  of  a  lateral  edge  by  the  perimeter  of.  the 
right  section. 


A 


E' 


Ji 


Let  GHIKL  be  a  right  section  of  the  prism  AU. 

To  prove  lateral  area  of  AD' =  AA\GH-\-  III+  etc.). 

Proof.  Consider  the  lateral  edges  AA\  BB\  etc.,  to  be  the 
bases  of  the  HI  AB\  BC\  etc.,  which  form  the  lateral  surface 
of  the  prism. 

Then  the  bases  of  these  UJ  are  all  equal.  §  551 

Since  the  sides  of  the  right  section,  GH,  HI,  etc.,  are  X  to 
AA\  BB',  etc.  (§  462),  they  are  the  altitudes  of  these  ZI7,  and 
the  sum  of  the  altitudes  OS",  HI^  IK,  etc.,  is  the  perimeter 
of  the  right  section. 

The  area  of  each  O  is  the  product  of  its  base  and  alti- 
tude. §  365 

Hence,  the  sum  of  the  areas  of  the  /17  is  the  product  of  a 
lateral  edge  AA^  by  the  perimeter  of  the  right  section. 

But  the  sum  of  the  areas  of  the  JJ]  is  the  lateral  area  of 
the  prism. 

Therefore  the  lateral  area  of  the  prism  is  equal  to  the  product 
of  a  lateral  edge  by  the  perimeter  of  a  right  section.         q.  e.  d. 

562.  Cor.  The  lateral  area  of  a  right  prism  is  equal  lo  the 
altitude  multiplied  by  the  perimeter  of  the  base. 


282 


SOLID   GEOMETRY. 


BOOK   VII. 


Proposition  III.     Theorem. 

563.  Two  prisms  are  equal  if  three  faces  including 
a  trihedral  angle  of  the  one  are  respectively  equal  to 
three  faces  including  a  trihedral  angle  of  the  other, 
and  are  similarly  placed. 

J  f 


In  the  prisms  AI  and  AT,  let  AD,  AG,  AJ,  be  respec- 
tively equal  to  A'D',  A'G',  A' J',  and  similarly  placed. 

To  prove  prism  AI=^ prism.  A^I'. 

Proof.   By  hypothesis  the  face  A  BAE,  BAF,  EAF,  are 
equal  to  the  face  A  B'A'E',  B'A'F',  F'A'F',  respectively. 
Therefore  the  trihedral  angle  A=A'.  §  641 

Apply  A  to  its  equal  A' ;  then  the  faces  AD,  AG,  A  J, 
will  coincide  with  the  equal  faces  A'B',  A'G',  A'J^,  respec- 
tively, the  points  Cand  D  falling  at  C  and  D'. 

As  the  lateral  edges  of  the  prisms  are  parallel,  CIST  will  take 
the  direction  of  O^E^,  and  DIoi  D'F. 

Since  the  points  F,  G,  and  J  coincide  with  F',  G',  and  J', 
each  to  each,  the  planes  of  the  upper  bases  will  coincide. 
Hence  ^will  coincide  with  EC',  and  7"  with  J'. 
Therefore  the  prisms  coincide  and  are  equal.       q.e.d. 

564.  Cor.  1.  Too  truncated  prisms  are  equal  if  three  faces 
including  a  trihedral  of  the  one  are  respectively  equal  to  three 
faces  including  a  trihedral  of  the  other,  and  are  similarly  placed, 

565.  Cor.  2.  Two  right  prisms  having  equal  bases  and  alti- 
tudes are  equal.  If  the  faces  are  not  similarly,  placed,  one  of  the 
prisms  can  be  inverted  and  applied  to  the  other. 


A 


PEISMS. 


283 


Proposition  IV.     Theorem. 

566.  An  oblique  -prism  is  equivalent  to  a  right  prism 
whose  base  is  equal  to  a  right  section  of  the  oblique 
prisma,  and  whose  altitude  is  equal  to  a  lateral  edge 
of  the  oblique  prism. 


Let  FI  be  a  right  section  of  the  oblique  prism  AD'. 

Produce  AA'  to  F',  making  FF'=  AA',  and  at  F'  pass  the 
plane  F'F  ±  to  FF',  cutting  all  the  faces  of  AZ>'  produced, 
and  forming  the  right  section  F'l'  equal  and  parallel  to  FI. 

To  prove  AD'^FF. 

Proof.     In  the  solids'  AI  and  A'F,  AD=  A'D'.  §  551 

Also  AG  =  A'G']  for,  AF=  A'F',  and  BG  =  B'G',  since 
AA'  =  FF'  and  BB'=  GG';  and  AB  and  FG  are  equal  and 
parallel  to  A'B'  and  F'G'  respectively,  since  AB'  and  FG'  are 
parallelograms  (§  551).  Therefore  J.  6^  and  A'G'  are  mutually- 
equilateral  and  equiangular,  and  hence  equal.  §  203 

In  like  manner  we  may  prove  ^-^and  B'^'  equal. 

Hence  the  truncated  prisms  ^/and  A' I'  are  equal.    §  564 

Taking  each  in  turn  from  the  whole  solid,  we  have 

AD'^FF.  ^^.D. 


y 


284  SOLID    GEOMETRY. BOOK    VII. 


Proposition  V.     Theorem. 

567.  Any  two  o-pposite  faces  of  a  parallelopiped  are 
equal  and  parallel, 

E  H 


1 


Let  AG  he  a,  parsbUelopiped. 

To  prove  faces  AF  and  DO  equal  and  'parallel. 

Proof.         ^^is  II  to  Z)(7and  ^^is  II  to  D^  §§658,168 

Hence                     Z  EAB  -  Z  HDC.  §  498 

Also                 AB  =  .DC^TidiAE=DS.     ^  §179 

Therefore  the  face  AF==  face  DG.  §  185 

Moreover,  the  face  ^i^is  II  to  DG,  §  498 

{if  two  A  not  in  the  same  plane  have  their  sides  II  and  lying  in  the  same 
direction,  their  planes  are  parallel). 

In  like  manner  any  two  opposite  faces  may  be  proved  equal 
and  parallel.  q.  e.  d. 

568.  Scholium.  Any  two  opposite  faces  of  a  parallelepiped 
may  be  taken  for  baaes,  since  they  are  equal  arfd  parallel 
parallelograms. 

Ex.  490.  Show  that  any  lateral  edge  of  a  right  prism  is  equal  to  the 
altitude. 

Ex.  491.    Show  that  the  lateral  faces  of  right  prisms  are  rectangles. 

Ex.  492.  Prove  that  every  section  of  a  prism  made  by  a  plane  par- 
allel to  the  lateral  edges  is  a  parallelogram. 


PRISMS. 


285 


Proposition  VI.     Theorem. 

569.  The  plane  passed  through  two  diagonally 
opposite  edges  of  a  parallelopiped  divides  the  paral- 
lelopiped  into  two  equivalent  triangular  prisms. 

G 


Let  the  plane  AEG C  pass  through  the  opposite  edges 
AE  and  CG  of  the  parallelopiped  AG. 

To  prove  that  the  parallelopiped  AG  is  divided  into  two 
equivalent  triangular  prisms  ABC-F  and  ACD-S. 

Proof.    Let  IJKL  be  a  right  section  of  the  parallelopiped 
made  by  a  plane  JL  to  the  edge  AE. 

Since  the  opposite  faces  are  parallel,  §  567 

IJ'-is.  II  to  LK,  and  IL  to  JK.  §  492 

Therefore  IJKL  is  a  parallelogram.  §  168 

The  intersection  IK  of  the  right  section  with  the  plane 

AEQG\%  the  diagonal  of  the  O  IJKL. 

:.AIKJ=AIKL.  §178 

But  the  prism  ABC-F  i^  equivalent  to  a  right  prism  whose 

base  is  /t/lfiTand  whose  altitude  is  AE^  and  the  prism  ACD-H 

is  equivalent  to  a  right  prism  whose  base  is  ILK,  and  whose 

altitude  is  AE.  §  566 

But  these  two  right  prisms  are  equal.  §  565 

.-.  ABC-F-  ACD-K.  o.  e.  a 


286 


SOLID    GEOMETEY. BOOK:    VII. 


Proposition  VII.     Theorem. 

570.  Two  rectangular  parallelopipeds  having  equal 
bases  are  to  each  other  as  their  altitudes. 


B 

/        p        / 

/                 / 

B- 

/               P'              / 

/                 / 

/                                 / 

/                 / 

/                                 / 

/                 / 

— 

— 

/^                             / 

/^                / 

7 


Let  AB  and  A'B'  be  the  altitudes  of  the  two  rectan- 
gular parallelopipeds  P  and  P\  having  equal  bases. 

To  prove  P  \  P^  =  AB  :  A^BK 

Case  I.   When  AB  and  AJB^  are  commensurable. 

Proof.    Find  a  common  measure  of  AB  and  A^B\ 

Suppose  this  common  measure  to  be  contained  in  AB  m 
times,  and  in  A^B^  n  times  ;  then  we  have 

AB  :  A^B'  =-m:n.  (1) 

At  the  several  points  of  division  on  AB  and  A'B^  pass 
planes  JL  to  these  lines. 

The  parallelepiped  P  will  he  divided  into  m, 

and  P^  into  n,  parallelopipeds,  equal  each  to  each.  §  565 

Therefore  P\P^  =  m\n.  (2) 

From  (1)  and  (2),  P :  P^  ^  AB  :  A'B'. 


PRISMS. 


287 


Case  II.    When  AB  and  A'B'  are  incommensurable. 


^ 


s 


p' 


s 


\ 


s 


\J 


Let  AB  he  divided  into  any  number  of  equal  parts, 

and  let  one  of  these  parts  be  applied  to  A'B'  as  a  unit  of 
measure  as  many  times  as  A'B'  will  contain  it. 

Since  AB  and  A'B'  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  from  A'  to  a  point  D,  leaving  a  re- 
mainder JDB'  less  than  one  of  the  parts. 

Through  I)  pass  a  plane  J.  to  A'B',  and  denote  the  parallel- 
epiped whose  base  is  the  same  as  that  of  F',  and  whose  altitude 
is  A'B,  by  Q. 

Now,  since  AB  and  A'B  are  commensurable, 

Q:F=A'I):AB.  Case  I. 

If  the  unit  of  measure  is  indefinitely  diminished,  these 
ratios  continue  equal,  and  they  approach  indefinitely  the  lim- 
iting ratios  F' :  F  and  A'B' :  AB  respectively. 

Therefore  F:F=  A'B' :  AB,  §  260 

{if  two  variables  are  constantly  equal,  and  each  approaches  a  limit,  their 
limits  are  equal). 

a  E.  D. 

571.  Scholium.  The  three  edges  of  a  rectangular  parallelo- 
piped  which  meet  at  a  common  vertex  are  its  ditnensions. 
Hence  two  rectangular  parallelopipeds  which  have  two  dimen- 
sions in  common  are  to  each  other  as  their  third  dimensions. 


288 


SOLID   aEOMETRY.  —  BOOK   VII. 


Proposition  VIII.     Theorem. 

572.  Two  rectangular  parallelopipeds  having  equal 
altitudes  are  to  each  other  as  their  bases» 


Let  a,  6,  and  c,  and  a\  h',  c,  be  the  three  dimensions 
respectively  of  the  two  rectangular  parallelopipeds 
P  and  P'. 

rp  P       axh 

10  prove  ~:;r.=—, r/ 

Let  Q  be  a  third  rectangular  parallelepiped  whose  dimen- 
sions are  a\  b,  and  c. 

Now  Q  has  the  two  dimensions  b  and  c  in  common  with  P, 
and  the  two  dimensions  a'  and  c  in  common  with  P'. 

Q      a'' 

{two  rectangular  parallelopipeds  which  have  two  dimensions  in  common  are 
to  each  other  as  their  third  dimensions). 

The  product  of  these  two  equalities  is 


Then 


and 


P 
P' 


ax 


Xb' 


aE.  o. 


573.  Scholium.  This  proposition  may  be  stated  as  follows  : 
Two  rectangular  parallelopipeds  which  have  one  dimension  in 
common  are  to  each  other  as  the  products  of  the  other  two 
dimensions. 


PRISMS. 


289 


Proposition  IX.     Theorem. 

574.  Two   rectangular  parallelopipeds  are  to  each 
other  as  the  products  of  their  three  dimensions. 


\ 


\ 

\  "'  ^ 

c 

s     \ 

c 

k      ' 

\      \ 

a' 


Let  a,  6,  c,  and  a',  6',  c',  be  the  three  dimensions  re- 
spectively of  the  two  rectangular  parallelopipeds  P 
and  P'. 

rn  F       axbxc 

Proof.  Let  Q  be  a  third  rectangular  parallelepiped  whose 
dimensions  are  a,  b,  and  c\ 

P^c_ 

Q     c'' 

Q  ^  axb 

P'      a'  X  b'' 

{two  rectangular  parallelopipeds  which  have  one  dimension  in  common 
are  to  each  other  as  the  products  of  the  other  two  dimensions). 

The  product  of  these  equalities  is 

_, ^   r-^  ^       aXbXc  ^^ 


Then 


and 


§571 
§573 


/^e-.T, 


t'  ><  b'  X  c' 


1/ 


a. 


ly^ 


Ex.  493.  j^ind  the  ratio  of  two  rectangular  parallelepipeds  if  their 
altitudes  are  each  6  inches,  and  their  bases  5  inches  by  4  inches,  and  10 
inches  by  8  inches,  respectively. 

Ex.  494.  Find  the  ratio  of  two  rectangular  parallelopipeds,  if  their 
dimensions  are  3,  4,  5,  and  9.  8,  10,  respectively. 


290 


SOLID   GEOMETEY. — BOOK   VII. 


Peoposition  X.     Theorem. 

575.  The  volume  of  a  rectangular  parallelopiped  is 
equal  to  the  product  of  its  three  dimensions,  the  unit 
of  volume  being  a  cube  whose  edge  is  the  linear  uniL 


TZT 


/ 

/ 

/ 

f 

/    / 

/ 

/ 

1 

Let  a,  b,  and  c  be  the  three  dimensions  of  the  rec- 
tangular parallelopiped  P,  and  let  the  cube  U  be  the 
unit  of  volume. 

To  prove  that  the  volume  of  P=aXh  X  c. 

P     aXbx  c 


Proof. 


U     1x1x1 


aXbXc. 


F  . 


Since  Uis  the  unit  of  volume,  —  is  the  volume  of  P. 


§574 
§547 


Therefore  the  volume  oi  F=aXb  X  c.  a  e.  d. 

576.    Cor.  1.   The  volume  of  a  cube  is  the  cube  of  its  edge. 

bllt  Cor.  2.  The  product  axb  represents  the  area  of  base 
when  c  is  the  altitude  ;  hence  :  The  volume  of  a  rectangular 
parallelopiped  is  equal  to  the  product  of  its  base  by  its  altitude. 

578.  Scholium.  When  the  three  dimensions  of  the  rectan- 
gular parallelopiped  are  each  exactly  divisible  by  the  linear 
unit,  this  proposition  is  rendered  evident  by  dividing  the  solid 
into  cubes,  each  equal  to  the  unit  of  volume.  Thus,  if  the 
three  edges  which  meet  at  a  common  vertex  contain  the  linear 
unit  3,  4,  and  5  times  respectively,  planes  passed  through  the 
several  points  of  division  of  the  edges,  and  perpendicular  to 
them,  will  divide  the  solid  into  cubes,  each  equal  to  the  unit 
of  volume ;  and  there  will  evidently  be  3  X  4  X  5  of  these  cubes. 


PEISMS. 


291 


Proposition  XI.     Tseorem. 

679.  The  volume  of  any  parallelopiped  is  equal  to 
the  product  of  its  base  by  its  altitude. 


Let  n  denote  the  altitude  of  the  parallelopiped  AG. 
To  prove  that  the  volume  AG  —  A  BCD  X  jff". 

Proof.  Consider  ADHE  the  base  of  AG,  and  prolong  the 
lateral  edges  AB,  DC,  EF,  HG. 

The  right  parallelopiped  A^G\  determined  by  two  right  sec- 
tions A'D<WE\  B'C'G'F\  with  lateral  edge  A'B'  =  AB,  is 
equivalent  to  AG.  §  566 

Again,  consider  D'C^G'JP  the  base  of  A^G\  and  prolong 
the  lateral  edges  D'A\  C'B',  W E\  G'F. 

Then  the  parallelopiped  A^O,  determined  by  two  right  sec- 
tions A^B^NM,  KLOP,  with  lateral  edge  A^K=  D^A'  is 
equivalent  to  A'G'(^  566),  and  hence  to  AG. 

The  three  solids  have  a  common  altitude  ^  (§  494),  and 
equivalent  bases,  hr  ABCD=o=A'B'C'D\^Se>6),  and  A'B'CD' 

=  A^KZB'(n^^)- 

But  ^'0  is  a  rectangular  parallelopiped,  for  the  right  sec- 
tions A*]^,  KO,  are  rectangles,  since  the  opposite  faces  A}P, 
B^O,  are  ±  to  A'B'LK 

Hence       the  volume  A'O  =  A'B'LKx  IT.      §  577 
Therefore  the  volume  AG  =  ABCD  X  ^ 


Q.  E.  0. 


292 


SOLID   GEOMETRY. 


BOOK    VII. 


Proposition  XIL     Theorem. 

580.  The  volume  of  a  triangular  prism  is  equal  to 
the  product  of  its  base  by  its  altitude. 


Let  V  denote    the   volume,   B   the  base,  and  H  the 
altitude  of  the  triangular  prism  AEC-E', 

To  prove  V=  B  X  H. 

Proof.   Upon  the  edges  AU,  EC,  EE\  construct  the  paral- 
lelepiped .^^OZ)-^'. 

Then,  since  a  plane  passed  through  two  diagonally  opposite 
edges  of  a  parallelepiped  divides  it  into  two  equivalent  trian- 
gular prisms,  §  669 
AEC-E' -\AECB-EK 

Since  the  volume  of  any  parallelepiped  is  equal  to  the  prod- 
uct of  its  base  by  its  altitude, 

AECD-E*  =  AECD  x  H.  §  579 

But  AECD  =  2B.  §178 

.-.  V=i(2Bxir)  =  Bxir. 


Ex.  495.   Find  the  volume  of  a  right  triangular  prism,  if  its  height  is 
14  inches,  and  the  sides  of  the  base  are  6.  6,  and  5  inches. 


PRISMS. 


293 


Proposition  XIII.     Theorem. 

581.  The  volume  of  any  ■prisin  is  equal  to  the  prod- 
uct of  its  base  by  its  altitude. 


Let  V  denote  the  volume,  B  the  base,  and  H  the 
altitude  of  the  prism  DA'. 

To  prove  F=  B  X  IT. 

Proof.  Planes  passed  through  the  lateral  edge  AA',  and  the 
diagonals  AC,  AD  of  the  base,  will  divide  the  given  prism 
into  triangular  prisms. 

The  volume  of  each  triangular  prism  is  equal  to  the  product 
of  its  base  by  its  altitude  (§  580) ;  and  hence  the  sum  of  the 
volumes  of  the  triangular  prisms  is  equal  to  the  sum  of  their 
bases  multiplied  by  their  common  altitude. 

But  the  sum  of  the  triangular  prisms  is  equal  to  the  given 
prism,  and  the  sum  of  their  bases  is  equal  to  the  base  of  the 
given  prism.  Therefore  the  volume  of  the  given  prism  is 
equal  to  the  product  of  its  base  by  its  altitude. 

That  is,  V=Bxir.  aE.D. 

582.  Cor.  The  volumes  of  two  prisms  are  to  each  other  as  the 
products  of  their  bases  and  altitudes ;  prisms  having  equivalent 
bases  are  to  each  other  as  their  altitudes ;  prisms  having  equal 
altitudes  are  to  each  other  as  their  bases ;  prisms  having  equiv- 
alent  bases  and  equal  altitudes  are  equivalent. 


294 


SOLID   GEOMETRY.  —  BOOK   VII. 


Pyramids. 

683.   A  pyramid  is  a  polyhedron  of  wWch  one  face,  called 
the  base,  is  a  polygon,  and  the  other  faces, 
called  lateral  faces,  are  triangles  having  a 
common  vertex,  called   the  vertex   of  the 
pyramid. 

584.  The  intersections  of  the  lateral  faces 
are  called  the  lateral  edges  of  the  pyramid.        ^  ^- 

585.  The  sum  of  the  areas  of  the  lateral  faces  is  called  the 
lateral  area  of  the  pyramid. 

586i    The  altitude  of  a  pyramid  is  the  length  of  the  perpen- 
dicular let  fall  from  the  vertex  to  the  plane  of  the  base. 

587.  A  pyramid  is  called  triangular,  quadrangular,  etc., 
according  as  its  base  is  a  triangle,  quadrilateral,  etc. 

588.  A  triangular   pyramid,  having  four  faces,  is  called  a 
tetrahedron,  and  any  one  of  its  faces  can  be  taken  for  its  base. 

589.  A  pyramid  is  regular  if  its  base  is  a  regular  polygon 
whose  centre  coincides  with  the  foot  of  the 
perpendicular  let  fall  from  the  vertex  to  the 
base. 


590.  The  lateral  edges  of  a  regular  pyramid 
are  equal,  since  they,  cut  off  equal  distances  from 
the  foot  of  the  perpendicular  let  fall  from  the    Regular  Pyramid, 
vertex  to  the  base  (§  478).    Therefore  the  lateral  faces  are  equal 
isosceles  triangles. 

591.  The  slant  height  of  a  regular  pyramid  is  the  length  of 
the  perpendicular  from  the  vertex  to  the  base  of  any  one  of  its 
lateral  faces.  It  is  the  common  altitude  of  all  the  lateral  faces, 
and  bisects  the  base  of  the  lateral  face  in  which  it  is  drawn. 

592.  A  frustum  of  a  pyramid  is  the  portion  of  a  pyramid 


PYRAMIDS, 


295 


included  between  its  base  and  a  plane  parallel  to  the  base 
and  cutting  all  the  lateral  edges. 

593.   The  altitude  of  a  frustum  is  the  length  of 
the  perpendicular  between  the  planes  of  its  bases, 

694.    The  lateral  faces  of  a  frustum  of  a  regu-^ 
lar  pyramid  are  equal  trapezoids. 

595.   The  slant  height  of  the  frustum  of  a  regular  pyramid 
is  the  altitude  of  one  of  these  trapezoids. 


Proposition  XIV.    Theorem. 

696.  The  lateral  area  of  a  regular  pyramid  is  equal 
to  one-half  the  product  of  the  slant  height  by  the 
perimeter  of  its  base. 


Let  S  denote  the  lateral  area  at  the  regular  pyra- 
mid V-ABCDE,  and  VII  its  slant  height 
To  prove  that  8-=^  VII{AB  -\'B0-\-  etc.). 
Proof.  The  A  VAB,  VBO,  etc.,  are  equal  isosceles  A.  §590 
Theareaof  the  sum  of  these  A=|F^(^^+^C+etc.)  §  368 
But  the  sum  of  their  areas  equals  the  lateral  area  of  the 
pyramid.  ...^=|KS-(^5+£C+etc.).  aE.a 

597.  Cor.  1.  The  latcn^al  area  of  the  frustum  of  a  regular 
pyramid,  is  equal  to  one-half  the  sum  of  the  perimeters  of  the 
bases  multiplied  by  the  slant  height  of  the  frustum,  §  371 


296 


SOLID    GEOMETRY.  —  BOOK    VII. 


Proposition  XV.     Theorem. 
598.  If  a  pyramid  is  cut  hy  a  plane  parallel  to  its 


I.  Th^  edges  and  altitude   are    divided   propor- 
tionally ; 

II.  The  section  is  a  polygon  similar  to  the  hase. 

V 


Let  V-ABCDE  be  cut  "by  a  plane  parallel  to  its  base, 

intersecting  the  lateral  edges  in  a,  b,  c,  d,  e,  and  the 

altitude  in  o. 

rp  J  Va       Vb  Vo  - 

To  prove   I.  _:=^....^  — ; 

IL  The  section  abode  similar  to  the  base  ABODE. 
I.  Proof.    Suppose  a  plane  passed  througli  the  vertex  V  11  to 
the  base. 

Since  the  edges  and  the  altitude  are  intersected  by  three 
parallel  planes, 

§499 


Vo^ 

vd 


Va  ^  Vb 
VA      vb" 

II.  Since  the  sides  ab,  be,  etc.,  are  parallel  respectively  to 
AB,  BC,  etc.,  §  492 

the  A  abc,  bed,  etc.,  are  equal  respectively  to  the  A  ABC, 
BCD,  etc.  §  498 

Therefore  the  two  polygons  are  mutually  equiangular. 


PYEAMIDS.  297 

Also,  since  the  sides  of  the  section  are  parallel  to  the  corre- 
sponding sides  of  the  base, 

A  Vab,  Vbc,   etc.,   are   similar   respectively   to   A  VAB, 

VBC,  etc. 

•  A.  =  fZ^\  =  Il=fJ^\  =  ^  etc 
"  AB     \VBj     BO     \VC)      CD'      " 

Hence  the  polygons  have  their  homologous  sides  propor- 
tional ; 

Hence  section  abode  is  similar  to  the  base  ABODE.    §  319 

599.  Cor.  1.  Any  section  of  a  pyramid  parallel  to  its  base 
is  to  the  base  as  the  square  of  its  distance  from  the  vertex  is  to 
the  square  of  the  altitude  of  the  pyramid. 

Since        i^  =  ^i^V— .    .•.-^=-^.  §305 

VO     \VBj     AB         VO'     AB' 

But  _abcde_^;^^  g  3^g 

ABODE     aS 
abode         Vo 


' '  ABODE      vo' 

600.  Cor.  2.  If  two  pyramids  having  equal  altitudes  are  cut 
by  planes  parallel  to  their  bases,  and  at  equal  distances  from 
their  vertices^  the  sections  will  have  the  same  ratio  as  their  bases. 

-p  abode     _  Vo 

°''  ABODE~  vo' 

and  j±_c_ ^  _K_o_  «  ^gg 

A'BO'      v^i'  ^ 

But  Vo  =  V'o\  and  VO^Va. 

.  .  abode  :  ABODE  =  a'b'o' :  A'B'O^. 
Whence      abode  :  a'b'o'  =  ABODE :  A'B'O^.  §  298 

601.  Cor.  3.  If  two  pyramids  have  equal  altitudes  and 
equivalent  bases,  sections  made  by  planes  parallel  to  their  bases^ 
and  at  equal  distances  from  their  vertices,  are  equivalent. 


298 


SOLID   GEOMETRY.  —  BOOK   VII. 


Proposition  XVI.     Theorem. 

602.  Two  triangular  pyramids  having  equivalent 
hases  and  equal  altitudes  are  equivalent. 


Let  S-ABC  and  S'-A'B'C  have  equivalent  bases  situ- 
ated in  the  same  plane,  and  a  common  altitude. 

To  prove  8-ABC<>  jS'-A'B'O'. 

Proof.  If  the  pyramids  are  not  equivalent,  suppose  S-ABC 
the  greater.     Divide  the  common  altitude  into  n  equal  parts. 

Through  the  points  of  division  pass  planes  II  to  the  plane  of 
their  bases.  The  corresponding  sections  of  the  pyramids  are 
equivalent.  §  601 

On  the  base  of  S-ABC,  and  on  each  section,  as  lower  base, 
construct  a  prism  with  lateral  edges  equal  and  parallel  to  AB. 

Similarly,  construct  a  prism  on  each  section  of  S^-A'B^C^^ 
as  upper  base. 

The  sum  of  the  first  series  of  prisms  is  greater  than  S-ABC, 
and  the  sum  of  the  second  series  is  less  than  S-A'B'C;  there- 
fore the  difference  between  S-ABC  si,nd  S^-A^B'C  is  less  than 
the  difference  betvf  een  the  sums  of  these  two  series  of  prisms. 

Each  prism  in  S'-A'B^C  is  equivalent  to  the  prism  next 
above  it  in  S-ABC  (§  582).  Hence  the  difference  between 
the  two  series  of  prisms  is  the  lowest  prism  of  the  first  series. 
But  by  increasing  n  indefinitely  this  can  be  made  less  than 
any  assigned  volume,  however  small. 

Therefore  the  two  pyramids  cannot  difier  by  any  volume  how- 
ever small ;  therefore  the  pyramids  are  equivalent.  q.  e.  d. 


PYRAMIDS.  •      299 

Proposition  XVII.     Theorem. 

603.  The  volume  of  a  triangular  pyramid  is  equal 
to  one-third  of  the  product  of  its  base  and  altitude. 

^r--v.-- 


Let  V  denote  the  volume,  and  H  the  altitude,  of  the 
triangular  pyramid  S-ABC. 

To  prove  V=  J  ABCx  H. 

Proof.  On  the  base  ABC  construct  a  prism  ABC-8ED, 
having  its  lateral  edges  equal  and  parallel  to  SB. 

The   prism  will   be  composed  of  the  triangular  pyramid 
8-ABC2JCidi  the  quadrangular  pyramid  S-ACDE. 
Through  8 A  and  8D  pass  a  plane  8 AD. 

This  plane  divides  the  quadrangular  pyramid  into  the  two 

triangular  pyramids  8-ACD  and  8-AED,  which  have  the 

same  altitude  and  equal  bases.  §  178 

.-.  8-ACD  =o  8-AED.  §  602 

Now  the  pyramid  8-AED  may  be  regarded  as  having  E8D 
for  its  base  and  J.  for  its  vertex. 

.'.8-AED  0^8- ABO. 

Hence  the  three  pyramids  into  which  the  prism  ABC-8ED 
is  divided  are  equivalent ;  the  pyramid  8- ABC  is  equivalent 
to  one-third  of  the  prism. 

But  the  volume  of  the  prism  is  equal  to  the  product  of  its 

base  and  altitude.  §  580 

.•.V=\ABCxn.  Q.E.Q 


300 


SOLID    GEOMETRY.  —  BOOK    VII. 


Proposition  XVIII.     Theorem. 

6p4'.  The  volume  of  any  pyramid  is  equal  to  one^ 
third  the  product  of  its  base  and  altitude, 

S 


Let  V  denote  the  volume  of  the  pyramid  S-ABCDE, 

To  prove  F=  ^  ABODE  x  SO. 

Proof.  Through  the  edge  SI>,  and  the  diagonals  of  the  hase' 
DA,  DB,  pass  planes. 

These  divide  the  pyramid  into  triangular  pyramids,  whose 
bases  are  the  triangles  which  compose  the  base  of  the  pyramid, 
and  whose  common  altitude  is  the  altitude  80  of  the  pyramid. 

The  volume  of  the  given  pyramid  is  equal  to  the  sum  of  the 
volumes  of  the  triangular  pyramids. 

But  the  sum  of  the  volumes  of  the  triangular  pyramids  is 
equal  to  one-third  the  sum  of  their  bases  multiplied  by  their 
common  altitude.  §  603 

That  is,  V=- 1  ABODE  x  SO.  a  e.  d. 

605.  Cor.  The  volumes  of  two  pyramids  are  to  each  other  as 
the  products  of  their  bases  and  altitude^;  pyramids  having 
equivalent  bases  are  to  each  other  as  their  altitudes;  pyramids 
having  equal  altitudes  are  to  each  other  as  their  bases;  pyramids 
having  equivalent  bases  and  equal  altitudes  are  equivalent 

606i  Scholium.  The  volume  of  any  polyhedron  may  be 
found  by  dividing  it  into  pyramids,  computing  their  volumes 
separately,  and  finding  the  sum  of  their  volumes. 


PYRAMIDS. 


301 


Proposition  XIX.     Theorem. 

607,  The  volumes  of  two  teiTrahedrons,  having  a  tri- 
hedral angle  of  the  one  equal  to  a  trihedral  angle  of 
the  other,  are  to  each  other  as  the  products  of  the 

three  edges  of  these  trihedral  angles. 

& 


Let  V  and  V^  denote  the  volumes  of  the  two  tetra- 
hedrons S-ABC  and  S-A'B'C,  having  the  common  tri- 
hedral angle  S. 

To  prove  —  = 

Proof.   Draw  CD  and  CD'  ±  to  the  plane  SA'B', 
and  let  their  plane  intersect  SA'B'  in  SDD'. 

The  faces  SAB  and  SA'B'  may  be  taken  as  the  bases,  and 
CD,  CD'  as  the  altitudes,  of  the  triangular  pyramids  SAB-C 
and  SA'B'-C. 


SAB  X  CD        SAB  ..  CD 


§605 


SA'B'x  CD'     SA'B'     CD' 
{any  two  pyramids  are  to  each  other  as  the  products  of  their  bases  and 

altitudes). 

But  .^^=^4x^.  §374 


and 


SA'B'     SA'  X  SB' 
CD  _^SC 
CD'     SC'     ■ 
{being  homologous  sides  of  the  similar  ^  SBC  and  8D^C^). 
.    V _  SAxSBxSC 
' '  V     SA'  X  SB'  X  SO*' 


§319 


aE.  D. 


302 


SOLID    GEOMETRY. 


BOOK   VII. 


Proposition  XX.     Theorem. 

608.  The  frustum  of  a  triangular  pyramid  is  equiv- 
alent to  the  sum  of  three  pyramids  whose  common 
altitude  is  the  altitude  of  the  frustum  and  whose 
bases  are  the  lower  base,  the  upper  base,  and  a  mean 
proportional  between  the  two  bases  of  the  frustvMm. 


Let  B  and  b  denote  the  lower  and  upper  bases  of 
the  frustum  ABC-DEF,  and  H  its  altitude. 

Through  the  vertices  A,  U,  C  and  U,  I),  C  pass  planes 
dividing  the  frustum  into  three  pyramids. 

Now  the  pyramid  E-ABC  has  for  its  altitude  H,  the  alti- 
tude of  the  frustum,  and  for  its  base  B,  the  lower  base  of  the 
frustum. 

And  the  pyramid  C-EDF  has  for  its  altitude  H,  the  alti- 
tude of  the  frustum,  and  for  its  base  h,  the  upper  base  of  the 
frustum.     Hence  it  only  remains 

To  prove  E-ADC  equivalent  to.  a  pyramid,  having  for  its 
altitude  H,  and  for  its  base  -\/ B  X  h. 

Proof.  E-ABQ  2.T\di  E-ADO,  regarded  as  having  the  com- 
mon vertex  (7,  and  their  bases  in  the  same  plane  BD,  have  a 
common  altitude. 


PYRAMIDS.  303 

.-.  C-ABE  :  C-ADE  =  A  AEB  :  A  AEE,       §  605 
{pyramids  having  equal  altitudes  are  to  each  other  as  their  bases). 

Now  since  the  A  AEB  and  AED  have  a  common  altitude, 
{that  is,  the  altitude  of  the  trapezoid  ABED), 

we  have  A  AEB  :  A  AEB  =  AB  :  BE,  §  370 

.-.  C-ABE  :  C-ABE  =  AB  :  BE. 

That  is,  E-ABC :  E-ABC=  AB  :  BE. 

In  like  manner  E-ABC  and  E-BFC,  regarded  as  having 
the  common  vertex  E,  and  their  bases  in  the  same  plane  DC, 
have  a  common  altitude. 

.-.  E-ABC :  E-BFC ^  A  ABC :  A  BFC        §  605 

But  since  the  A  ABC  and  BFC  have  a  common  altitude, 
{that  is,  the  altitude  of  the  trapezoid  ACFD), 

we  have  A  ABC :  A  BFC  =  AC :  BE.  §370 

.-.  E-ABC :  E-BFC=  AC :  BF. 

But  -A  BEE  is  similar  to  A  ABC,  §  598 

{the  section  of  a  pyramid  made  by  a  plane  II  to  the  base  is  a  polygon 
similar  to  the  base). 

.\AB:BE=AC:BF.  §319 

.-.  E-ABC :  E-ABC  =  E-ABC :  E-BFC 

Now  E-ABC  ^ilTxB,  §603 

and  E-BFC  =  C-EBF=\Hx  b. 

.'.E-ABC=-ViE:xBxiirxb  =  iJI-VB><b. 
Hence,  E-ABC  is  equivalent  to  a  pyramid,  having  for  its 
altitude  E',  and  for  its  base  -\/B  X  b.  a  e.  d. 

609.  Cor.  If  the  volume  of  thefrustuyn  of  a  triangular  pyra- 
inid  is  denoted  by  V,  the  lowei'  base  by  B,  the  upper  base  by  6, 
and  the  altitude  by  H, 

V=\IIxB-^\Hxb^\nxVB^Xh 

=^\iix{B-\-b-{--\rB~^). 


/'-' 


/ 


SOLID   GEOMETRY.  - 


BOOK    VII. 


^  > 

J 


Pkoposition  XXI.     Theorem. 


j^ 


610.  T/z/g  voluTne  of  the  frustum  of  any  pyramid  is 
equal  to  the  sum  of  the  volumes  of  three  pyramids 
whose  common  altitude  is  the  altitude  of  the  frus- 
tum, and  whose  bases  are  the  lower  hase,  the  upper 
hase,  and  a  mean  proportional  between  the  bases  of 
the  frustum. 

S  T 


Let  B  and  b  denote  the  lower  and  upper  bases,  B 
the  altitude,  and  V  the  volume  of  ABCD-EFGL 

To  prove  V=\II{B  +  5  +  VB~xh). 

Proof.  Let  T-KLM  be  a  triangular  pyramid  having  the 
same  altitude  as  8-ABCD  and  its  base  KLM=^  ABCD,  and 
lying  in  the  same  plane.     Then  T-KLM ^  8-ABCD.    §  605 

Let  the  plane  UFG I  cut  T-KLM  in  NOP. 

Then  NOP  =o=  EFGL  §  601 

Hence  T-NOP=o=  8-EFGL 
Taking  away    the   upper   pyramids    leaves    the    frustums 
equivalent. 

But  the  volume  of  the  frustum  of  the  triangular  pyramid  is 

equal  to  ^ ^(^  +  ^  +  VB^b).  §  609 

.-.  v=  \b:{b  +  h  +  -VB^b).  Q.  E.  D. 


NUMERICAL    EXERCISES.  -'    ""*?         305^  v'  ' 


^ 


^ '  •  Numerical  Exercises. 


71-%  ' 


496.  Find  the  length  of  an  edge  of  a  cubical  vessel  which  will  hold 
2-ton8  of  water. 

497.  How  many  square  feet  of  lead  will  be  required  to  line  a  cistern, 
open  at  the  top,  which  is  4  feet  6  inches  long,  2  feet  8  inches  wide,  and 
contains  42  cubic  feet  ?      i  ^'T 

^498.   An  open  cistern  is  made  of  iron  2  inches  thick.     The  inner    o 
dimensions  are:   length,  4  feet  6  inches;  breadth,  3  feet ;  depth,  2  feet     '^ 
6  inches.     What  will  the  cistern  weigh  (i.)  when  empty  ?  (ii.)  when  full  ^ 
of  water?    Specific  gravity  of  iron  =- 7.2.    IfYxAAjf^  ^U/.«t  "  iT/  ^^ 

'  499.  An  open  cistern  6  feet  long  and  4J  feet  wide  holds  108  cubic  feet 
of  water.  How  many  cubic  feet  of  lead  will  it  take  to  line  the  sides  and 
bottom,  if  the  lead  is  \  inch  thick? 

500.  The  three  dimensions  of  a  rectangular  parallelopiped  are  a,  ?»,  c ; 
find  the  surface,  the  volume,  and  the  length  of  a  diagonal. 

501.  The  base  of  a  right  prism  is  a  rhombus,  one  side  of  which  is  10 
inches,  and  the  shorter  diagonal  is  12  inches.  The  height  of  the  prism 
is  15  inches.     Find  the  entire  surface  and  the  volume. 

502.  Find  the  volume  of  a  regular  hexagonal  prism  whose  height  is 
10  feet,  each  side  of  the  hexagon  being  10  inches. 

503.  A  pyramid  15  feet  high  has  a  base  containing  169  square  feet. 
At  what  distance  from  the  vertex  must  a  plane  be  passed  parallel  to  the 
base  so  that  the  section  may  contain  100  square  feet?    /?  '   ^ 

"  504.  The  base  of  a  pyramid  contains  144  square  feet.  A  plane  par- 
allel to  the  base  and  4  feet  from  the  vertex  cuts  a  section  containing  64 
square  feet ;  find  the  height  of  the  pyramid.    ' 

505.  A  pyramid  12  feet  high  has  a  square  base  measuring  8  feet  on  a 
side.  What  will  be  the  area  of  a  section  made  by  a  plane  parallel  to  the 
base  and  4  feet  from  the  vertex  ?     / 1  T 

506.  Two  pyramids  standing  on  the  same  plane  are  14  feet  high.     The  f\  = 
first  has  for  base  a  square  measuring  9  feet  on  a  side  ;  the  second  a  reg- 
ular hexagon  measuring  7  feet  on  a  side.    Find  the  areas  of  the  sections  ,  ^, 
made  by  a  plane  parallel  to  their  bases  and  6  feet  from  their  vertices.       'T**'' 

507.  The  base  of  a  regular  pyramid  is  a  hexagon  of  which  the  side 

measures  3  feet.     Find  the  height  of  the  pyramid  if  the  lateral  area  is 
equal  to  ten  times  the  area  of  the  base.  jj/ 


306  SOLID   GEOMETRY. — BOOK   VII. 


Proposition  XXII.    Theorem.  ' 

611.  A  truncated  triangular  prism  is  equivalent  to 
the  sum  of  three  pyramids  whose  common  base  is  the 
base  of  the  prism,  and  whose  vertices  are  the  three 
vertices  of  the  inclined  section. 


Let  ABC-DEF  be  a  truncated  triangular  prism  whose 
base  is  ABC,  and  inclined  section  DEF. 

Pass  the  planes  AEC  and  DEC,  dividing  the  truncated 
prism  into  the  three  pyramids  E-ABC,  E-ACD,  and  E-CBF. 
To  prove  ABC-DEF  equivalent  to  the  sum  of  the  three  pyra- 
mids, E-ABC,  D-ABC,  and  F-ABC 

Proof.   E-ABC ho.^  the  base  ^^Cand  the  vertex  E. 

The  pyramid  E-A  CD  =o  B-A  CD,  §  602 

(for  they  have  the  same  base  A  CD  and  the  same  altitude,  since  their  vertices 
E  and  B  are  in  the  line  EB  II  to  the  base  ACD). 

But  the  pyramid  B-ACD  may  be  regarded  as  having  the 
base  ABC  and  the  vertex  D ;  that  is,  as  D-ABC. 

The  pyramid  E-CDF<^  B-ACF, 
for  their  bases  CDF  and  ACF,  in  the  same  plane,  are  equiva- 
lent, §  369 

{since  the  A  CDF  and  A  CF  have  the  common  base  CF  and  equal  altitudes, 
their  vertices  lying  in  the  line  AD  II  to  CF), 


PYRAMIDS. 


307 


and  the  pyramids  liave  the  same  altitude, 

{since  their  vertices  E  and  B  are  in  the  line  EB  II  to  the  plane  of  their 

bases  ACDF). 

But  the  pyramid  B-ACF  may  be  regarded  as  having  the 
base  ^^Cand  the  vertex  F\  that  is,  as  F-ABC. 

Therefore  the  truncated  triangular  prism  ABC-DEF  is 
equivalent  to  the  sum  of  the  three  pyramids  F-ABC,  D-ABC, 
and  F-ABC.  a  e.  d 


612.  Cor.  1.  The  volume  of  a  truncated  right  triangular 
prism  is  equal  to  the  product  of  its  base  by  one-third  the  sum 
of  its  late)'al  edges.  For  the  lateral  edges  DA,  EB,  FC,  being 
perpendicular  to  the  base,  are  the  altitudes  of  the  three  pyra- 
mids whose  sum  is  equivalent  to  the  truncated  prism.  And, 
since  the  volume  of  a  pyramid  is  one-third  the  product  of  its 
base  by  its  altitude,  the  sum  of  the  volumes  of  these  pyramids 
=  ABGx\{DA-\-EB  +  FC). 

613.  Cor.  2.  The  volume  of  any  truncated  triangular  prism 
is  equal  to  the  p7'oduct  of  its  right  section  by  one-third  ^e  sum 
of  its  lateral  edges.  For  let  ABC-A'B'C  be  any  truncated 
triangular  prism.  Then  the  right  section  DEF  divides  it 
into  two  truncated  right  prisms  whose  volumes  are 

DEFx  J  {AD^  BE-^  CF)  and  DEFx  \  {A*I)-{-B'E-\-  C'F). 

Whence  their  sum  is  I)EFxi(AA'+  BB'-\-CC^). 


308 


SOLID   GEOMETRY. 


BOOK   VII. 


Similar  Polyhedrons. 

614.  Similar  polyhedrons  are  polyhedrons  that  have  the 
same  number  of  faces,  respectively  similar  and  similarly  placed, 
ar^  their  corresponding  polyhedral  angles  equal. 

Homologous  faces,  lines,  and  angles  of  similar  polyhedrons 
are  faces,  lines,  and  angles  similarly  placed. 

615.  Cor.  1.  The  homologous  edges  of  similar  polyhedrons 
are  proportional.  §  319 

616.  CoR.  2.  Two  homologous  faces  of  similar  polyhedrons  are 
proportional  to  the  squares  of  two  homologous  edges.  §  377 

617.  CoR.  3.  The  entire  surfaces  of  two  similar  polyhedrons 
are  proportional  to  the  squares  of  two  homologous  edges.    §  303 

618.  Cor.  4.  The  homologous  dihedral  angles  of  similar 
polyhedrons  are  equal. 


Proposition  XXIII.    Theorem. 

619.  Two  similar  polyhedrons  may  he  decomposed 
into  the  same  number  of  tetrahedrons  similar,  each 
to  eojch,  and  similarly  placed. 


Let  P  and  P'  be  two  similar  polyhedrons. 


SIMILAR  POLYHEDRONS.  309 

To  prove  that  the  similar  polyhedrons  P  and  P'  can  he 

decomposed  into  the  same  number  of  tetrahedrons,  similar  each 
to  each,  and  similarly  placed. 

Proof.  Through  the  vertices  A,  G,  C,  and  the  homologous 
vertices  A\  G',  0',  pass  planes. 

The  tetrahedrons  G-ABC  and  G'-A'B'C  have  the  faces 
ABC,  GAB,  GBC,  similar  respectively  to  A'B'C\  G'A'B', 
G'B'C.  §  332 

Hence  in  the  faces  G^^Cand  G'A'C 

AG  ^(  AB\_  AC  _f  BC\^  GO         §  oiq 
A'G'     \A'B'J     A'C     \B'cO     G'C'        ^ 

Therefore  the  face  G AC  is  similar  to  G'A'C.      §  324 

Hence  the  faces  of  these  tetrahedrons  are  similar,  each  to 
each. 

Also,  any  two  corresponding  trihedral  A  of  these  tetrahe- 
drons are  equal.  §  541 

Therefore  the  tetrahedron  G-ABC  is  similar  to  G'-A'B'C. 

§614 

If  G-ABC  and  G'-A'B'C  be  removed,  the  polyhedrons 
remaining  will  continue  similar  ;  for  the  new  faces  GAC  and 
G'A'C  have  just  been  proved  similar,  and  the  modified  faces 
^6^i^ and  A'G'F*,  CGJI &ud  C'G'IT',  will  be  similar  (§  332) ; 
also  the  modified  polyhedral  A  G  and  G',  A  and  A\  (7 and 
C,  will  remain  equal  each  to  each,  since  the  corresponding 
parts  taken  from  them  are  equal. 

The  process  of  removing  similar  tetrahedrons  can  be  carried 
on  until  the  polyhedrons  are  reduced  to  tetrahedrons ;  that  is, 
until  the  two  similar  polyhedrons  are  decomposed  into  the 
same  number  of  tetrahedrons  similar  each  to  each,  and  simi- 
larly situated. 

a  E.  D. 

620.  Cor.  Any  two  homologous  lines  in  two  similar  polyhe- 
drons have  the  same  ratio  as  any  two  homologous  edges. 


310 


SOLID   GEOMETRY.  —  BOOK   VII. 


Proposition  XXIV.    Theorem. 

621.  The  volumes  of  two  similar  tetrahedrons  are 
to  each  other  as  the  cubes  of  their  homologous  edges. 


Let  V  and  V  denote  the  volumes  of  the  two  similar 
tetrahedrons  S-ABC  and  S'-A'B'C, 


To  prove 


V_^  SB" 


Proof.   Since  the  homologous  trihedral  angles  8  and  >S"  are 
equal,  we  have 

V         SBxSCxSA 


V     jS'B'  X  S'C  X  jS'A' 

^  SB       SO       SA 
S'B'     S'C     S'A'' 


But 


SB       SO 


V 

v 


S'B' 
SB 


S'O' 
xMr.X 


SA 
S'A'' 

SB 


SR 


S'B'     S'B'     S'B'     -^T^'' 


§607 


§615 


Q.  E.  D. 


Ex.  508.  The  homologous  edges  of  two  similar  tetrahedrons  are  as 
6  :  7.     Find  the  rat^  tf  the'i/  s&faces  and  of  tij/ei^ volumes. 

Ex.  509.  If  the  edge  of  a  tetrahedron  is  a,  find  the  homologous  edge 
of  a  similar  tetrahedron  twice  as  large.    ^^  ■  ^  ^ 


/7^-  "ira}  :  o, :  s'<rL 


SIMILAR   POLYHEDRONS. 


311 


Proposition  XXV.     Theorem. 


622.  The  volumes  of  two  similar  polyhedrons  are  to 
each  other  as  the  cubes  of  any  two  homologous  edges 
L 


Let  r,  V'  denote  the  volumes,  QB,  G'B'  any  two 
homologous  edges,  of  the  polyhedrons  P  and  P*. 

To  prove  V:V'=0£'  :  GHf. 

Proof.  Decompose  these  polyhedrons  into  tetrahedrons  sim- 
ilar, each  to  each,  and  similarly  placed.  §  619 

Denote  the  volumes  of  these  tetrahedrons  by  v,  Vj,  Vj, , 

v\  Vi,  V2', Then 


OB' 


Vi 


GS      V,  _0F 


v'       QiB''     v^      Q'£t'     v^      QijBf^ 


§621 


—  z=—Lz=z  -!l. 

v'       Vi        V2 


Whence 


or 


t;  -f-  ^1  +  ^2   _  '^  _  GrB 
v'  +  Vi'  +  V,'     v'      qFb^' 

V  _0B' 


§303 


Q.B.D 


312  SOLID    GEOMETRY. —  BOOK   VII. 


Regular  Polyhedrons. 

623.  A  regular  polyhedron  is  a  polyhedron  whose  faces  are 
equal  regular  polygons,  and  whose  polyhedral  angles  are  equal. 

Proposition  XXVI.     Problem. 

624.  To  determine  the  numher  of  regular  convex 
polyhedrons  possible. 

A  convex  polyhedral  angle  must  have  at  least  three  faces, 
and  the  sum  of  its  face  angles  must  be  less  than  360°  (§  640). 

1.  Since  each  angle  of  an  equilateral  triangle  is  60°,  convex 
polyhedral  angles  may  be  formed  by  combining  three,  four,  or 
five  equilateral  triangles.  The  sum  of  six  such  angles  is  360°, 
and  therefore  greater  than  the  sum  of  the  face  angles  of  a  convex 
polyhedral  angle.  Hence  not  more  than  three  regular  convex 
polyhedrons  are  possible  with  equilateral  triangles  for  faces. 

2.  Since  each  angle  of  a  square  is  90°,  a  convex  polyhedral 
angle  may  be  formed  by  combining  three  squares.  The  sum 
of  four  such  angles  is  360°,  and  therefore  greater  than  the 
sum  of  the  face  angles  of  a  convex  polyhedral  angle.  Hence 
only  one  regular  convex  polyhedron  is  possible  with  squares. 

3.  Since  each  angle  of  a  regular  pentagon  is  108°,  a  convex 
polyhedral  angle  may  be  formed  by  combining  three  regular 
pentagons.  The  sum  of  four  such  angles  is  432°,  and  there- 
fore greater  than  the  sum  of  the  face  angles  of  a  convex  poly- 
hedral angle.  Hence  only  one  regular  convex  polyhedron  is 
possible  with  regular  pentagons. 

4.  We  can  proceed  no  further  ;  for  the  sum  of  three  angles 
of  regular  hexagons  is  360°,  of  regular  heptagons  is  greater 
than  360°,  etc.  Hence  only  five  regular  convex  polyhedrons 
are  possible. 

There  are  five  regular  polyhedrons  called,  from  the  number 
of  faces,  the  tetrahedron,  the  hexahedron^  the  octahedron,  the 
dodecahedron,  the  icosahedron. 


REGULAR   POLYHEDRONS. 


313 


Proposition  XXVII.     Problem. 

625.   Upon  a  given  edge  to  construet  tJie  regular 
polyhedrons. 


Let  AB  be  the  given  edge. 

Upon  AB  to  construct  the  regular  polyhedrons. 

1.  Construction  of  the  Eegular  Tetrahedron.  Upon  the  given 
edge  construct  an  equilateral  triangle.  At  its  centre  erect 
a  X  to  its  plane,  and  take  a  point  D  in  this  ±  such  that  DA 
=  AB.  Join  D  to  each  of  the  vertices  of  the  triangle  ABC. 
The  polyhedrotf  D-ABCD  is  a  regular  tetrahedron. 

Proof.  The  four  faces  are  by  construction  equal  equilateral 
triangles  (§  480),  and  the  four  trihedral  angles  A,  B,  C,  D,  are 
equal,  since  their  face  angles  are  all  equal.  §  541 

Therefore  D-ABC  is  a  regular  tetrahedron. 

2.  Construction  of  the  Eegular  Hexahedron.   Upon  the  given 

edge  AB  construct  the  square  ABCD,  and       jr ^Qt 

upon  the  sides  of  this  square  construct  the        /  1^ 

squares  AF,  BO,  CH,  BE,  L  to  the  plane    ^ 
ABCD. 

The  polyhedron  ^6^  is  a  regular  hexa- 
hedron. 

Proof.   The  six  faces  are  by  construction    ^ 
equal  squares,  and  the  eight  trihedral  angles  A,  B,  C,  D,  E, 
F,  O,  II,  are  equal  since  their  face  angles  are  all  equal.  §  541 

Therefore  AQ\s,z,  regular  hexahedron. 


I 


314  SOLID  GEOMETRY.  —  BOOK   VII. 

3.  Construction  of  the  Eegular  Octahedron.  Upon  the  given 
edge  AB  construct  the  square  ABCD^ 

and  through  its  centre  0  pass  a  X  to  its  J^ 

plane.  y^  W  - 

On  this  JL  take  the  points  E  and  F      /       /L\p  \. 
such  that  AE  and  AF  are  each  equal  C'^^^^^^i^J^^^^^^^k^ 

to  AB.  ^rB\^/^^^ 

Join  ^and  i^to  each  of  the  vertices  \,\7  y^ 

of  the  square  ABCD.     The  polyhedron  \|^^ 

E-ABOD-F  is  a  regular  octahedron. 

Proof.  Since  all  the  lines  from  E  and  F  to  A,  B,  C,  and  D 
are  equal  (§  480),  and  each  equal  to  AB,  the  eight  triangles 
which  form  the  faces  are  equal  and  equilateral. 

Since  O  is  the  centre  of  the  square  ABCD,  the  diagonal  of 
this  square  ^C' will  pass  through  0,  and  the  lines  ^i^and  AC 
which  intersect  in  0  are  in  the  same  plane.  Hence  E,  C,  F, 
and  A  are  in  one  plane. 

In  the  A  AEG,  ABQ  AFC,  the  side  AC  is  common,  and 
all  the  other  sides  equal.  Therefore  these  triangles  are  equal 
(§  160);  and  since  Z  ABO  is  a  right  angle,  AEO  and  AFO 
are  right  angles. 

Therefore  AECF  is  a  square  equal  to  the  square  ABOD. 
Hence  the  pyramid  B-AECF  has  its  four  faces  and  its  base 
^^Ci^  equal  to  the  four  faces  and  the  base  ABCD  of  the 
pyramid  E-ABCD. 

Therefore  the  two  pyramids  are  equal,  and  the  tetrahedral 
angle  B  is  equal  to  the  tetrahedral  angle  E. 

In  like  manner  it  can  be  shown  that  any  other  two  polyhe- 
dral angles  are  equal.  Therefore  the  polyhedron  is  a  regular 
octahedron. 

4.  Construction  of  the  Eegular  Dodecahedron.  Construct  a  reg- 
nlaff  pentagon  M  with  its  sides  equal  each  to  the  given  edge, 
and  join  to  each  of  its  sides  the  side  of  an  equal  pentagon  so 
inclined  to  the  plane  of  M  as  to  form  trihedral  angles  at  its 


REGULAR   POLYHEDRONS. 


315 


vertices.  Construct  a  regular  pentagon  J/'  =  M,  and  join  to 
each  of  its  sides  the  side  of  an  equal  pentagon  so  inclined 
to  the  plane  of  Jf'  as 
to  form  trihedral  an- 
gles at  its  vertices. 

We  now  have  two 
equal  convex  surfaces 
composed  each  of  six 
equal  regular  penta- 
gons. The  trihedral 
angles  formed  at  the  vertices  of  M  and  Jf'  are  equal,  each  to 
each  (§  541) ;  therefore  the  dihedral  angles  are  all  equal,  and 
the  two  surfaces  can  be  combined  so  as  to  form  a  single  convex 
surface. 

Proof.  Put  the  two  surfaces  together  with  their  convexities 
turned  in  opposite  directions,  so  that  the  vertex  a  and  the  side 
ah  shall  coincide  with  the  vertex  B  and  the  side  BA  respec- 
tively. Then  two  consecutive  face  angles  of  one  surface  will 
unite  with  a  single  face  angle  of  the  other,  and  form  a  trihe- 
dral angle,  since  any  two  consecutive  faces  contain  a  dihedral 
angle  of  one  of  the  trihedral  angles  already  formed  at  the  ver- 
tices of  M  and  M\  The  trihedral  angles,  therefore,  are  all 
equal,  and  the  polyhedron  is  a  regular  dodecahedron. 

5.  Oonstruction  of  the  Eegular  Icosahedron.  Construct  a  regu- 
lar pentagon  ABODE,  with  its  sides  equal 
each  to  the  given  edge.  At  its  centre  erect 
a  X  to  its  plane,  and  in  this  perpendicu- 
lar take  a  point  such  that  PA  =  AB.  Join 
P  with  each  of  the  vertices  of  the  pentagon,  forming  a  regular 
pentagonal  pyramid,  whose  vertex  is  P,  and  whose  dihedral 
angles  formed  cy^  the  edges  PA,  PB,  etc.,  are  all  equal.  §  542 

Complete  the  pentahedral  angles  at  A,  B,  C,  etc.,  adding 
to  each  three  equilateral  triangles  each  equal  to  PAB,  and 
making  the  dihedral  angles  about  A,  B,  O,  etc.,  all  equal. 


E^- 


316 


SOLID    GEOMETRY.  —  BOOK    VII. 


Construct  a  regular  pentagonal   pyramid   P-A^B'Q^DE^ 
equal  to  P-ABCDE.     This  can  be  joined 
to  the  convex  surface  already  formed,  so  as 
to  form  a  single  convex  surface. 

Proof.  Two  consecutive  face  angles  of 
one  surface  will  unite  with  three  consecu- 
tive face  angles  of  the  other,  and  form  a 
regular  pentahedral  angle,  since  they  have 
together  three  dihedral  angles  of  such  a  pentahedral  angle. 

The  pentahedral  angles  are  therefore  all  equal,  and  the 
polyhedron  is  a  regular  icosahedron. 

626.  Scholium.  The  regular  polyhearons  may  be  con- 
structed as  follows : 

Draw  the  diagrams  given  below  on  cardboard.  Cut  through 
the  full  lines  and  half  through  the  dotted  lines.  Bring  the 
edges  together  so  as  to  form  the  respective  polyhedrons,  and 
keep  the  edges  in  contact  by  pasting  along  them  strips  of 
strong  paper. 


Tetrahedron. 


Hexahedron. 


Octahedron. 


Dodecahedron. 


Icosahedron, 


POLYHEDRONS.  317 

General  Theorems  of  Polyhedrons. 

Proposition  XXVIII.    Theorem.    (Euler's.) 

627.  In  any  polyhedron  the  number  of  edges  in- 
creased hy  two  is  equal  to  the  number  of  vertices 
increased  hy  the  number  of  faces. 


Let  E  denote  the  number  of  edges,  V  the  number  of 
vertices,  F  the  number  of  faces,  of  S-ABCDE. 

To  prove  U-\-2=V+F, 

Proof.   Beginning  with  one  face  ABODE,  we  have  E=V. 

Annex  a  second  face  SAB,  by  applying  one  of  its  edges  to 
a  corresponding  edge  of  the  first  face,  and  there  is  formed  a 
surface  having  one  edge  AB  and  two  vertices  A  and  B  com- 
mon  to  the  two  faces. 

Therefore,  for  two  faces  E=  V-{- 1. 

Annex  a  third  face  8B0,  adjoining  each  of  the  first  two  faces; 
this  face  will  have  two  edges,  SB,  BC,  and  three  vertices  S,  B, 
C,  in  common  with  the  surface  already  formed. 

Therefore,  for  three  faces    E=  F+  2. 

In  like  manner,         for  four  faces      E=  V-}-  3. 

And  so  on  for  (E-l)  faces  E=V+  (F-  2), 

But  E~l  is  the  number  of  faces  of  the  polyhedron  when 

only  one  face  is  lacking,  and  the  addition  of  this  face  will  not 

increase  the  number  of  edges  or  vertices.     Hence,  for  E  faces 

E=V+F-2,  or  E+2  =  V+F.  0.1.^ 


318  SOLID   GEOMETRY.  —  BOOK    VII. 


Proposition  XXIX.     Theorem. 

628.  The  sum  of  the  face  angles  of  any  polyhedron 
is  equal  to  four  right  angles  taken  as  many  times, 
less  two,  as  the  polyhedron  has  vertices. 

S 


B  c 

Let  E  denote  the  number  of  edges,  V  the  number  oi 
vertices,  F  the  number  of  faces,  and  S  the  sum  of  the 
face  angles,  of  the  polyhedron  S-ABCDE, 

To  prove  8  -=  (F—  2)  4  rt.  A. 

Proof.  Since  ^  denotes  the  number  of  edges,  2  E  will  denote 
the  number  of  sides  of  the  faces,  considered  as  independent 
polygons,  for  each  edge  is  common  to  two  polygons.' 

If  an  exterior  angle  is  formed  at  each  vertex  of  every  poly- 
gon, the  sum  of  the  interior  and  exterior  angles  at  each  vertex 
is  2  rt.  ^ ;  and  since  there  are  2  E  vertices,  the  sum  of  the 
interior  and  exterior  angles  of  all  the  faces  is 
2^x2rt.J,  or  ^x4rt.  A 

But  the  sum  of  the  ext.  A  of  each  face  is  4  rt.  A  (§  207), 
and  the  number  of  faces  is  F;  therefore  the  sum  of  all  the 
ext.  Ah  i^X  4  rt.  A. 

Therefore  8,  the  sum  of  the  int.  A,  is 
{E-  F)  4  rt.  A. 

But  ^+  2  =  r+  i^(§  627)  ;  that  is,  E-F=V-  2. 

Therefore  8=-  (Y—  2)  4  rt.  A. 

aB.D 


CYLINDERS. 


319 


The  Cylinder. 


629.  A  cylindrical  surface  is  a  curved  surface  generated  by 
a  moving  straight  line  AB,  called  the  generatrix,  which  moves 
parallel  to  itself  and  constantly  touches  a 
fixed  curve  BODE,  called  the  directrix. 
The  generatrix  in  any  position  is  called 
an  element  of  the  surface.  One  element, 
and  only  one,  can  be  drawn  through  a 
given  point  of  a  cylindrical  surface. 

630.  A  cylinder  is  a  solid  bounded  by 
a  cylindrical  surface  and  two  parallel  planes  which  cut  all  the 
elements.  The  two  plane  surfaces  are  called  the  bases,  and  the 
cylindrical  surface  is  called  the  lateral  surface. 

631.  The  altitude  of  a  cylinder  is  the  length  of  the  perpen- 
dicular between  the  planes  of  its  bases.  The  elements  of  a 
cylinder  are  all  equal. 

632.  A  right  section  of  a  cylinder  is  a  section  made  by  a 
plane  perpendicular  to  its  elements. 

633.  A  cylinder  is  a  right  cylinder  if  its  elements  are  per- 
pendicular to  its  bases  ;  otherwise  it  is  an  oblique  cylinder. 

634.  A  circular  cylinder  is  a  cylinder  whose  base  is  a  circle. 

635.  A  cylinder  of  revolution  is  a  cylinder  generated  by  the 
revolution  of  a  rectangle  about  one  side  as  an 
axis. 


636.  Similar  cylinders  of  revolution  are  cyl- 
inders generated  by  similar  rectangles  revolv- 
ing about  homologous  sides. 

637.  A  tangent  line  to  a  cylinder  is  a  straight  line,  not  an 
element,  which  touches  the  surface  of  the  cylinder  but  does 
not  intersect  it. 


320  SOLID   GEOMETRY.  —  BOOK   VII. 

638i  A  plane  whicli  contains  an  element  of  tlie  cylinder  and 
does  not  cut  the  surface,  is  called  a  tangent  plane.  The  ele- 
ment contained  by  the  plane  is  called  the  element  of  contact. 

639t  A  prism  is  inscribed  in  a  cylinder  when  its  lateral 
edges  are  elements  of  the  cylinder  and  its  bases  are  inscribed 
in  the  bases  of  the  cylinder. 

640.  A  prism  is  circumscribed  about  a  cylinder  when  its 
lateral  edges  are  parallel  to  elements  of  the  cylinder  and  its 
bases  are  circumscribed  about  the  bases  of  the  cylinder. 

Proposition  XXX.     Theorem. 

641.  Every  section  of  a  cylinder  made  hy  a  plane 
passing  through  an  element  is  a  parallelogram. 


Let  a  plane  pass  through  the  element  AD  of  the 
cylinder  AC. 

To  prove  the  section  A  BCD  a  parallelogram. 

Proof.  A  plane  passing  through  the  element  AD  will  cut 
the  circumference  of  the  base  in  a  second  point  B. 

The  straight  line  BC  drawn  II  to  AD  lies  in  the  plane  DAB 
(§  98) ;  and  it  is  an  element  of  the  cylinder.  §  629 

Hence  BO  ia  the  intersection  of  the  plane  and  the  surface 

of  the  cylinder.     Also  DO  is  II  to  AB.  §  492 

Therefore  ABCD  is  a  parallelogram.  §  168 

642,  Cor.  Every  section  of  a  right  cylinder  made  by  a  plane 
passing  through  an  element  is  a  rectangle. 


CYLINDERS. 


321 


Proposition  XXXI.     Theorem. 

i.  The  hoses  of  a  cylinder  are  equal, 
O^ _c 


Let  ABE  and  DCG  be  the  bases  of  the  cylinder  AC, 

To  prove  ABU  =  DCG. 

Proof.   Let  A,  B^  E,  be  any  three  points  in  the  perimeter  of 

the  lower  base,  and  AD^  BC^  EG,  be  elements  of  the  surface. 

Join  AE,  AB,  EB,  DG,  DC,  GO. 

Then  AQAG,  EC  are  £17.  §  182 

.-.  AE=  BG,  AB  =  BC,  and  EB  =  GO.        §  179 

.•.AABE=ABCG.  §160 

Apply  the  upper  base  to  the  lower  base  so  that  BC,  shall 

fall  upon  AB. 

Then  G  will  fall  upon  E. 

But  G  is  any  point  in  the  perimeter  of  the  upper  base, 

therefore  every  point  in  the  perimeter  of  the  upper  base  will 

fall  upon  the  perimeter  of  the  lower  base. 

Therefore  the  bases  coincide  and  are  equal.- 

aE.D. 

644.  Cor.  1.  Any  two  parallel  sections  ABO  and  A*B'0\ 
cutting  all  ike  elements  of  a  cylinder  EF,  are  equal.  For  these 
sections  are  the  bases  of  the  cylinder  A  C\ 

645.  Cor.  2.  Ani/  section  of  a  cylinder  parallel  to  the  base  is 
equal  to  the  base. 


322  SOLID   GEOMETRY. — BOOK   VII. 


Proposition  XXXII.    Theorem. 

646.  The  lateral  area  of  a  cylinder  is  equal  to  the 
product  of  the  perimeter  of  a  right  section  of  the 
cylinder  hy  an  element  of  the  surface. 


Let  S  denote  the  lateral  area,,  P  the  perimeter  of  a 
right  section,  and  E  an  element  of  the  surface  of  AC\ 

To  prove  8=PxE. 

Proof.    Inscribe  in  the  cylinder  a  prism  having  for  its  base 

the   polygon  ABODE,  and  denote  the  lateral   area  of  this 

prism  by  5,  and  the  perimeter  of  the  right  section  ahcde  by  p. 

Then  s=pxE.  §  561 

Let  the  number  of  lateral  faces  of  the  inscribed  prism  be 
indefinitely  increased,  the  new  edges  continually  dividing  the 
arcs  in  the  bases  of  the  cylinder.  Then  the  perimeters  of  the 
bases  of  the  prism  will  approach  the  perimeters  of  the  bases 
of  the  cylinder  as  limits,  and  the  lateral  area  of  the  prism 
will  approach  the  lateral  area  of  the  cylinder  as  a  limit.  Hence 
the  perimeter  of  the  right  section  of  the  prism  will  approach 
the  perimeter  of  the  right  section  of  the  cylinder  as  a  limit. 

But,  however  great  the  number  of  faces,  s=pX  E. 

.'.S=FxE.  §260 

Q.  E.  D. 

647.  Cor.  1.  The  lateral  area  of  a  cylinder  of  revolution  is 
the  product  of  the  circumference  of  its  base  hy  its  altitude. 


CYLINDERS. 


323 


648.  Cor.  2.  If  8  denotes  the  lateral  area,  T  the  total  area,  H 
the  altitude,  and  R  the  radius,  of  a  cylinder  of  revolution, 
8=2ttBxII. 
T=  2TrR  X  H+  2TrR'  =  2'jrR{H-\-  R). 


Proposition  XXXIII.     Theorem. 

649.  The  volume  of  a  cylinder  is  equal  to  the  prod- 
uct of  its  base  hy  its  altitude. 

.A 


Let  V  denote  the  volume,  B  the  base,  and  H  the 
altitude,  of  the  cylinder  AG. 

To  prove  V=  BxH. 

Proof.  Let  V*  denote  the  volume  of  the  inscribed  prism  AG, 
and  -B'  its  base.     The  altitude  of  this  prism  will  be  H. 

Then  V=B'xS.  §581 

If  the  number  of  lateral  faces  of  the  inscribed  prism  is  in- 
definitely increased,  the  new  edges  continually  dividing  the 
arcs  of  the  bases,  B*  approaches  -B  as  a  limit,  and  V*  approaches 
Fas  its  limit. 

But  however  great  the  number  of  the  lateral  faces, 

V'=B'xH.     :.V  =  BxB:.  §260 

ae.o. 

650.  Cor.  If  V  denotes  the  volume,  R  the  radium,  H  the 
altitude,  of  a  cylinder  of  revolution,  then  the  area  of  the  base 
is  ttR",  and  V=  irR'  X  H.  . 


324 


SOLID   GEOMETEY.  —  BOOK   VII. 


Proposition  XXXIV.     Theorem. 

651.  The  lateral  areas,  or  the  total  areas,  of  siTuilar 
cylinders  of  revolution  are  to  each  other  as  the  squares 
of  their  altitudes,  or  of  their  radii;  and  their  volumes 
are  to  each  other  as  the  cubes  of  their  altitudes,  or 
of  their  radii. 


Let  S,  S'  denote  the  lateral  areas,  T,  T'  the  total 
areas,  F,  F  the  volumes,  H,  H*  the  altitudes,  R,  R'  the 
radii,  of  two  similar  cylinders  of  revolution. 

Toprove  8  :  8'=^T   :  T'  =^  IP :  W' ^  B'  :  B'\ 
and  V:V'  =  ir':  IT"  =  B'  :  B'\ 

Proof.   Since  the  generating  rectangles  are  similar, 
H _R  _  H+R 

Therefore,  by  §§  648,  650, 
8 


§§  319,  303 


8' 


"IttRH  _R       H _^  R^__  H^ 
2'jrR'W     R'     W     R''     H'^' 


^     27rR' (JI'+ R')     r\jI'+R'J       R" 


JT 


jj;n 


V  ^-^R'H  ^R\  H  _R^  ^IP 

V  ttR^H'.    R''     H'  ~R"     H^^ 


Q.E.O. 


CONES. 


325 


The  Cone. 

652.  A  conical  surface  is  the  surface  generated  by  a  moving 
straight  line  called  the  generatrix,  passing  through  a  fixed 
point  called  the  vertex,  and  constantly  touching  a  fixed  curve 
called  the  directrix. 


653.  The  generatrix  in  any  position  is 
called  an  element  of  the  surface.  If  the 
generatrix  is  of  indefinite  length,  the  surface 
consists  of  two  portions,  one  above  and  the 
other  below  the  vertex,  which  are  called  the 
upper  and  lower  nappes,  respectively. 

Through  a  given  point  in  a  conical  surface 
one  element,  and  only  one,  can  be  drawn. 


654.  If  the  directrix  is  a  closed  curve,  the 

solid  bounded  by  the  conical  surface  and  a  plane  cutting  all 
its  elements  is  called  a  cone.  The  conical  surface  is  called  the 
lateral  surface,  and  the  plane  surface  the  hose,  of  the  cone. 
The  length  of  the  perpendicular  from  the  vertex  to  the  plane 
of  the  base  is  called  the  altitude  of  the  cone. 

655.  A  circular  cone  is  a  cone  whose  base  is  a  circle.  The 
straight  line  joining  the  vertex  and  the  centre  of  the  base  is 
called  the  axis  of  the  cone. 

If  the  axis  is  perpendicular  to  the  base,  the  cone  is  called  a 
right  cone ;  otherwise,  the  cone  is  called  an  oblique  cone. 

656.  A  right  circular  cone  is  a  cone  whose  axis  is  perpen- 
dicular to  its  base,  and  is  called  a  cone  of  revolu- 
tion, because  it  may  be  generated  by  the  revolu- 
tion of  a  right  triangle  about  one  of  its  legs  as 
an  axis.  The  hypotenuse  in  any  position  is  an 
element  of  the  surface,  and  is  called  the  slant 
height  of  the  cone. 


326  SOLID   GEOMETRY.  —  BOOK   VII. 

657.  Similar  cones  of  revolution  are  cones  generated  by  the 
revolution  of  similar  right  triangles  about  homologous  legs. 

658.  A  tangent  line  to  a  cone  is  a  line,  not  an  element, 
which  touches  the  surface  of  the  cone  and  does  not  cut  it. 

659.  A  plane  which  contains  an  element  of  the  cone  and 
does  not  cut  the  surface,  is  called  a  tangent  plane.  The  ele- 
ment contained  by  the  plane  is  called  the  element  of  contact. 

660.  A  pyramid  is  inscribed  in  a  cone  when  its  lateral  edges 
are  elements  of  the  cone  and  its  base  is  inscribed  in  the  base 
of  the  cone. 

661.  A  pyramid  is  circumscribed  about  a  cone  when  its 
base  is  circumscribed  about  the  base  of  the  cone  and  its  vertex 
coincides  with  the  vertex  of  the  cone. 

662.  A  frustum  of  a  cone  is  the  portion  of  a  cone  included 
between  the  base  and  a  section  parallel  to  the  base  and  cut- 
ting all  the  elements. 

663.  The  base  of  the  cone  is  called  the  lower  base  of  the 
frustum,  and  the  parallel  section  the  upper 
base. 


664.  The  altitude  of  a  frustum  of  a  cone  is 
the  length  of  the  perpendicular  between  the 
planes  of  its  bases. 

665.  The  lateral  surface  of  a  frustum  of  a  cone  is  the  por- 
tion of  the  lateral  surface  of  the  cone  included  between  the 
bases  of  the  frustum. 

666.  The  slant  height  of  a  frustum  of  a  cone  of  revolution  is 
the  portion  of  any  element  of  the  cone  included  between  the 
bases. 


CONES. 


Proposition  XXXV.     The 


667.  Every  section  of  a  cone  made  hy  a  plane  pass- 
ing through  its  vertex  is  a  triangle. 


Let  a  plane  pass  through  the  vertex  S  and  cut  the 
base  in  BD. 

To  prove  the  section  8BD  a  triangle. 

Proof.         Draw  the  straight  lines  8B  and  8D. 

Then  8B  and  8D  are  elements  of  the  surface  of  the  cone, 
and  they  lie  in  the  cutting  plane,  since  they  have  each  two 
points  in  common  with  the  plane.  Hence  they  are  the  inter- 
sections of  the  conical  surface  with  the  cutting  plane. 


And  BD  is  a  straight  line. 
Therefore  the  section  8BD  is  a  triangle. 


§471 


Q.E.D 


Ex.  510.  Show  that  any  lateral  face  of  a  pyramid  circumscribed  about 
a  cone  is  tangent  to  the  cone. 

Ex.  511.   The  diagonals  of  a  parallelepiped  bisect  each  other, 

Ex.  512.   The  square  of  a  diagonal  of  a  rectangular  parallelepiped  is 
equal  to  the  sum  of  the  squares  of  its  three  dimensions.  ^^ 


328  SOLID   GEOMETRY.  —  BOOK   VII. 


Proposition  XXXVI.     Theorem. 

668.  Every  section  of  a  circular  cone  made  hy  a 
plane  parallel  to  the  base  is  a  circle* 


Let  the  section  ahc  of  the  circular  cone  S-ABC  be 
parallel  to  the  base. 

To  prove  that  abc  is  a  circle. 

Proof.   Let  0  be  the  centre  of  the  base,  and  let  o  be  the  point 
in  which  the  axis  SO  pierces  the  plane  of  the  parallel  section. 
Through  SO  and  the  elements  SA,  SB,  etc.,  pass  planes 
cutting  the  base  in  the  radii  OA,  OB,  etc., 

and  the  section  ahc  in  the  straight  lines  oa,  oh,  etc. 
Since  ahc  is  II  to  ABC,  oa  and  oh  are  tl  respectively  to  OA 
and  OB.  §  492 

Therefore  the  A  Soa  and  Sob  are  similar  respectively  to  the 

§§  106,  321 
h\^_oh_ 
Vj      OB 

OB.  §  211 

.*.  oa  =  oh. 
That  is,  all  the  straight  lines  drawn  from  o  to  the  perimeter 
of  the  section  are  equal. 

.'.  the  section  ahc  is  a  O.  q.  e.  ot 

669.    Cor.   The  axis  of  a  circular  cone  passes  through  the 
centres  of  all  the  sections  which  are  parallel  to  the  base. 


A  SOA  and  SOB. 

.    oa  _ 
"OA 

But 

OA 

CONES. 


329 


Proposition  XXXVII.     Theorem. 

670.  The  lateral  area  of  a  cone  of  revolution  is  equal 
to  one-half  the  product  of  the  slant  height  hy  the  cir- 
cumference of  the  base. 


Let  S  denote  the  lateral  area,  C  the  circtimference 
of  the  base,  and  L  the  slant  height,  of  the  cone. 

To  prove  S^^iCxL. 

Proof.  Circumscribe  about  the  base  any  regular  polygon 
ABCD,  and  upon  this  polygon  as  a  base  construct  the  regu- 
lar pyramid  S-ABCD  circumscribed  about  the  cone. 

If  the  lateral  area  of  this  pyramid  is  s,  the  perimeter  p,  the 
slant  height  L,s  =  \pX  L.  §  596 

Let  the  number  of  the  lateral  faces  of  the  circumscribed 
pyramid  be  indefinitely  increased,  the  new  edges  continually 
bisecting  the  arcs  of  the  base.  Then  p  and  s  approach  C  and 
8  respectively  as  their  limits. 

But  however  great  the  number  of  lateral  faces  of  the 
pyramid,  s  =  ^pxL. 

.'.S=iCxL.  §260 


671.  Cor.  Since  C=2^E, 

S  =  i(2TrBX  L)  =  irRL. 

The  total  area      T=i^RL^-tTR^  =itR(^L-\-R\ 


aE.  D. 
§419 


330 


SOLID   GEOMETRY. BOOK    VII. 


Proposition  XXXVIII.     Theorem. 

672.  The  volume  of  any  cone  is  equal  to  the  product 
of  one-third  of  its  hase  hy  its  altitude. 


Let  V  denote  the  volume,  B  the  base,  and  H  the 
altitude  of  the  cone. 

To  prove  V=\BxII. 

Proof.  Let  the  volume  of  an  inscribed  pyramid  A-CDEFG 
be  denoted  by  V\  its  base  by  B^  and  its  altitude  by  H. 

Then  V'  =  \B'XII.  §604 

Let  the  number  of  lateral  faces  of  the  inscribed  pyramid 
be  indefinitely  increased,  the  new  edges  continually  dividing 
the  arcs  in  the  base  of  the  cone.  Then  F'  approaches  "p^as  its 
limit,  and  B^  approaches  B  as  its  limit. 

But  however  great  the  number  of  lateral  faces  of  the  pyramid, 

V^  =  \B^XR. 
.-.V  =\B  xH.  §260 

Q.  E.  D. 

673.  Cor.  If  the  cone  is  a  cone  of  revolution,  and  R  is  the 
radius  of  the  base,  j^  ~  ^^2  /g  425), 

and  V^IttE'x  H. 


CONES. 


331 


Proposition  XXXIX.     Theorem. 

674.  The  lateral  areas,  or  the  total  areas,  of  two 
similar  cones  of  revolution  are  to  each  other  as  the 
squares  of  their  altitudes,  or  of  their  radii;  and 
their  volumes  are  to  each  other  as  the  cubes  of  their 
altitudes,  or  of  tlzeir  radii. 


Let  S  and  S'  denote  the  lateral  areas,  T  and  T'  the 
total  areas,  V  and  V  the  volumes,  H  and  W  the  alti- 
tudes, R  and  R'  the  radii,  L  and  U  the  slant  heights, 
of  two  similar  cones  of  revolution. 


Toprove  S  :  S' =  T   :  T'   =  H' :![''==  H' :  H''- 
and  V:  V'=  IP  :  B:"  ==  E'  :  B"  =  B  :  n\ 

Proof.    Since  the  generating  triangles  are  similar, 
H _R  _L  _  L-\-R 

Therefore,  by  §§  671,  673, 

S      irRL      R  ^^L      R'      II      H^ 


L''.V\ 


§§  319,  303 


^R_     L  ^^ 

irR^n'    R'     n     R^' 


8' 

irRjL+R) 


V       irR\r-\-R') 


=  -^x 


L  +  R 


R'     L'i-R' 


R" 


JTrR'H  _I^  y  H  _R'  ^H' 
^ttR'^IT      R"     H'      R"     IT" 


aE.1 


332 


SOLID   GEOMETEY.  —  BOOK  VII. 


Proposition  XL.     Theorem. 

675.  The  lateral  area  of  the  frustum  of  a  cone  of 
revolution  is  equal  to  one-half  the  sum  of  the  circum- 
ferences of  its  bases  multiplied  hy  the  slant  height. 


Let  S  denote  the  lateral  area,  C  and  c  the  circum- 
ferences of  its  bases,  R  and  r  their  radii,  and  L  the 
slant  height.  ^ 

To  prove  8=-^{Q-{- c)x  L. 

Circumscribe  about  tbe  frustum  of  the  cone  the  frustum  of 
the  regular  pyramid  ABCD-A^B^C'D\  and  denote  the  lateral 
area  of  this  frustum  by  s,  the  perimeters  of  its  lower  and  upper 
bases  by  P  and  p  respectively,  and  its  slant  height  by  L. 

Then  s=^^{P+p)xL.  §  597 

Let  the  number  of  lateral  faces  be  indefinitely  increased,  the 
new  elements  constantly  bisecting  the  arcs  of  the  bases.  Then 
P  and  p  approach  C  and  c,  respectively,  as  their  limits. 

But,  however  great  the  number  of  lateral  faces  of  the  frus- 
tum of  the  pyramid, 

s  =  \{P^p)xL.     .'.S=i(C+c)xL.        §260 

aE.  D. 

676.  Cor.  The  lateral  area  of  a  frustum  of  a  cone  of  revo- 
lution is  equal  to  the  circumference  of  a  section  equidistant  from 
its  bases  multiplied  hy  its  slant  height. 


CONES.  333 


Proposition  XLI.     Theorem. 

677.  The  volume  of  a  ftiAystum  of  a  cone  is  equivo/- 
lent  to  the  sum  of  the  volumes  of  three  cones  whose 
common  altitude  is  the  altitude  of  the  frustum  and 
whose  bases  are  the  lower  base,  the  upper  base,  and  a 
mean  proportional  between  the  bases  of  the  frustum. 


Let  V  denote  the  volume  of  the  frustum.,  B  its  lower 
base,  b  its  upper  base,  and  H  its  altitude. 

To  prove  V=  i  II{B  +  i  +  VB><^b). 

Proof.  Let  V*  denote  the  volume,  £'  and  b^  the  lower  and 
upper  bases,  and  -ff"the  altitude,  of  an  inscribed  frustum  of  a 
pyramid. 

Then  r  =  i  B-(B'  +  b'-\-  V^'  X  b').  §610 

Let  the  number  of  lateral  faces  of  the  inscribed  frustum  be 
indefinitely  increased,  the  new  edges  continually  dividing  the 
arcs  in  the  bases  of  the  frustum  of  the  cone.  Then,  however 
great  the  number  of  lateral  faces  of  the  frustum  of  the 
pyramid.  yf^;^  ^^^r  _^  j,  _^  VBxI'y 

.'.V  =-iB:(B  '\-b-\-VBxb).  §  260 

a  E.  D. 

678.  Cor.  If  the  frustum  is  that  of  a  cone  of  revolution,  and 
R  and  r  are  the  radii  of  its  bases,  we  have  B  =  tt^',  b  =  ttt^, 
and  ^B xb  —  irRr. 

...  F=  4  irHiE'  +  r'  +  Rr). 


334    "^  SOLID    GEOMETRY.  —  BOOK   VII. 

Numerical  Exercises. 
The  Pyramid. 

Find  the  volume  in  cubie  feet  of  a  regular  pyramid  : 

513.  When  its  base  is  a  square,  each  side  measuring  3  feet  4  inches, 
and  its  height  is  9  feet. 

514.  When  its  base  is  an  equilateral  triangle,  each  side  measuring 
4  feet,  and  its  height  is  15  feet. 

515.  When  its  base  is  a  regular  hexagon,  each  side  measuring  6  feet, 
and  its  height  is  30  feet. 

Find  the  total  surface  in  square  feet  of  a  regular  pyramid : 

516.  When  each  side  of  its  square  base  is  8  feet,  and  the  slant  height 
is  20  feet. 

517.  When  each  side  of  its  triangular  base  is  6  feet,  and  the  slant 
height  is  18  feet. 

518.  When  each  side  of  its  square  base  is  26  feet,  and  the  perpendic- 
ular height  is  84  feet. 

Find  the  height  in  feet  of  a  pyramid  when  : 

519.  The  volume  is  26  cubic  feet  936  cubic  inches,  and  each  side  of  its 
square  base  is  3  feet  6  inches. 

520.  The  volume  is  20  cubic  feet,  and  the  sides  of  its  trls-ngular  base 
are  5  feet,  4  feet,  and  3  feet. 

521.  The  base  edge  of  a  regular  pyramid  with  a  square  base  meas- 
ures 40  feet,  the  lateral  edge  101  feet ;  find  its  volume  in  cubic  feet. 

522.  Find  the  volume  of  a  regular  pyramid  whose  slant  height  is  12 
feet,  and  whose  base  is  an  equilateral  triangle  inscribed  in  a  circle 
having  a  radius  of  10  feet. 

523.  Having  given  the  base  edge  a,  and  the  total  surface  T,  of  a 
regular  pyramid  with  a  square  base,  find  the  volume  V. 

524.  The  base  edge  of  a  regular  pyramid  whose  base  is  a  square  is  a, 
the  total  surface  T\  find  the  height  of  the  pyramid. 

525.  The  eight  edges  of  a  regular  pyramid  with  a  square  base  are 
equal  in  length,  and  the  total  surface  is  T;  find  the  length  of  one  edge. 

526.  Find  the  base  edge  a  of  a  regular  pyramid  with  a  square  base, 
having  given  the  height  h  and  the  total  surface  T. 


NUMERICAL   EXERCISES.  335 


Cylinders  and  Cones. 

^  527.  If  the  total  surface  of  a  right  circular  cylinder  closed  at  both 
ends  is  o,  and  the  radius  of  the  base  is  r,  what  is  the  height  of  the 
cylinder  ? 

^  528.  If  the  lateral  surface  of  a  right  circular  cylinder  is  a,  and  the 
volume  is  b,  find  the  radius  of  the  base  and  the  height. 

{  529.  How  many  cubic  yards  of  earth  must  be  removed  in  construct- 
ing a  tunnel  100  yards  long,  whose  section  is  a  semicircle  with  a  radius 
of  10  feet  ? 

530.  If  the  diameter  of  a  well  is  7  feet,  and  the  water  is  10  feet  deep, 
how  many  gallons  of  water  are  there,  reckoning  7J  gallons  to  the  cubic 
foot? 

*"  531.  When  a  body  is  placed  under  water  in  a  right  circular  cylinder 
60  centimeters  in  diameter,  the  level  of  the  water  rises  30  centimeters ; 
find  the  volume  of  the  body. 

^532.  If  the  circumference  of  the  base  of  a  right  circular  cylinder  is 
c,  and  the  height  h,  find  the  volume  V. 

533.  Having  given  the  total  surface  T  of  a  right  circular  cylinder, 
in  which  the  height  is  equal  to  the  diameter  of  the  base,  find  the 
volume  V. 

^  534.  If  the  circumference  of  the  base  of  a  right  circular  cylinder  is 
c,  and  thedotal  surface  is  T,  find  the  volume  V. 

^  535.   The  slant  height  of  a  right  circular  cone  is  2  feet.    At  what 

distance   from   the   vertex  must  the  slant  height  be  cut  by  a  plane 

parallel  to  the  base,  in  order  that  the  lateral  surface  may  be  divided 

into  two  equivalent  parts  ? 
y^  536.   The  height  of  a  right  circular  cone  is  equal  to  the  diameter  of 

its  base ;  find  the  ratio  of  the  area  of  the  base  to  the  lateral  surface. 
>»-^  587.   "What  length  of  canvas  f  of  a  yard  wide  is  required  to  make  a 

conical  tent  12  feet  in  diameter  and  8  feet  high  ? 
^  538.   The  circumference  of  the  base  of  a  circular  cone  is  12^  feet,  and 

its  height  8\  feet ;  find  its  volume. 
/^  539.   Given   the   total  surface  T  of  a  right  circular   cone,  and  the 

radius  r  of  the  base  ;  find  the  volume  V. 
^>^540.  Given   the  total  surface  T  of   a.  right  circular   cone,  and  the 

lateral  surface  8;  find  the  volume  V. 


336  SOLID    GEOMETRY.  —  BOOK   VII. 


Frustums  of  Pyramids  and  Cones. 

^^,---^541.  How  many  square  feet  of  tin  will  be  required  to  make  a  funnel 
if  the  diameters  of  the  top  and  bottom  are  to  be  28  inches  and  14  inches 
respectively,  and  the  height  24  inches  ? 

542.  Find  the  expense  of  polishing  the  curved  surface  of  a  marble 
column  in  the  shape  of  the  frustum  of  a  right  cone  whose  slant  height 
is  12  feet,  and  the  radii  of  the  circular  ends  are  3  feet  6  inches  and  2 
feet  4  inches  respectively,  at  60  cents  a  square  foot. 

"^43.  The  slant  height  of  the  frustum  of  a  regular  square  pyramid  is 
20  feet,  the  length  of  each  side  of  its  base  40  feet,  of  each  side  of  its  top 
16  feet ;  find  its  volume. 

^  544.  If  the  bases  of  the  frustum  of  a  pyramid  are  two  regular  hexa- 
gons whose  sides  are  1  foot  and  2  feet  respectively,  and  the  volume  of 
the  frustum  is  12  cubic  feet :  find  its  height. 

545.  The  frustum  of  a  right  circular  cone  is  14  feet  high,  and  has  a 
volume  of  924  cubic  feet.  Find  the  radii  of  its  bases  if  their  sum  is  9  feet. 
-  546.  From  a  right  circular  cone  whose  slant  height  is  30  feet,  and 
circumference  of  whose  base  is  10  feet,  there  is  cut  off  by  a  plane  parallel 
to  the  base  a  cone  whose  slant  height  is  6  feet.  Find  the  convex  surface 
and  the  volume  of  the  frustum. 

\  547.  Find  the  difference  between  the  volume  of  the  frustum  of  a 
pyramid  whose  bases  are  squares,  measuring  8  feet  and  6  feet  respec- 
tively on  a  side,  and  the  volume  of  a  prism  of  the  same  altitude  whose 
base  is  a  section  of  the  frustum  parallel  to  its  bases  and  equidistant  from 
them. 

•  548.  A  Dutch  windmill  in  the  shape  of  the  frustum  of  a  right  cone  is 
12  meters  high.  The  outer  diameters  at  the  bottom  and  the  top  are  16 
meters  and  12  meters,  the  inner  diameters  12  meters  and  10  meters, 
respectively.  How  many  cubic  meters  of  stone  were  required  to 
build  it? 

549.  The  chimney  of  a  factory  has  the  shape  of  a  frustum  of  a  regu- 
lar pyramid.  Its  height  is  180  feet,  and  its  upper  and  lower  bases  are 
squares  whose  sides  are  10  feet  and  16  feet  respectively.  The  flue  is 
throughout  a  square  whose  side  is  7  feet.  How  many  cubic  feet  of 
material  does  the  chimney  contain  ? 
P"^  550.  Find  the  volume  V  of  the  frustum  of  a  cone  of  revolution, 
^    having  given  the  slant  height  a,  the  height  \  and  the  convex  surface  S. 


NUMEEICAL   EXERCISES.  337 


Equivalent  Solids. 

551.  A  cube  whose  edge  is  12  inches  long  is  transformed  into  a  right 
prism  whose  base  is  a  rectangle  16  inches  long  and  12  inches  wide. 
Find  the  height  of  the  prism,  and  the  difference  between  its  total  surface 
and  the  surface  of  the  cube. 

^  552.  The  dimensions  of  a  rectangular  parallelopiped  are  a,  6,  c 
Find  (i.)  the  height  of  an  equivalent  right  circular  cylinder  having  a 
for  the  radius  of  its  base ;  (ii.)  the  height  of  an  equivalent  right  circular 
cone  having  a  for  the  radius  of  its  base. 

'^  553.  A  regular  pyramid  12  feet  high  is  transformed  into  a  regular 
prism  with  an  equivalent  base ;  what  is  the  height  of  the  prism  ? 

^  554.  The  diameter  of  a  cylinder  is  14  feet,  and  its  height  is  8  feet; 
find  the  height  of  an  equivalent  right  prism,  the  base  of  which  is  a 
square  with  a  side  4  feet  long. 

^  555.  If  one  edge  of  a  cube  is  a,  what  is  the  height  h  of  an  equivalent 
right  circular  cylinder  whose  diameter  is  5  ? 

""  556.  The  heights  of  two  equivalent  right  circular  cylinders  are  as 
4:  9.  The  diameter  of  the  first  is  6  feet;  what  is  the  diameter  of  the 
other? 

^  557.  A  right  circular  cylinder  6  feet  in  diameter  is  equivalent  to  a 
right  circular  cone  7  feet  in  diameter.  If  the  height  of  the  cone  is  8 
feet,  what  is  the  height  of  the  cylinder  ? 

'^558.  The  frustum  of  a  regular  four-sided  pyramid  is  6  feet  high,  and 
the  sides  of  its  bases  are  5  feet  and  8  feet  respectively.  What  is  the 
height  of  an  equivalent  regular  pyramid  whose  base  is  a  square  with  a 
side  12  feet  long  ? 

^  559.  The  frustum  of  a  cone  of  revolution  is  5  feet  high,  and  the 
diameters  of  its  bases  are  2  feet  and  3  feet  respectively ;  find  the  height 
of  an  equivalent  right  circular  cylinder  whose  base  is  equal  in  area  to 
the  section  of  the  frustum  made  by  a  plane  parallel  to  its  bases,  and 
equidistant  from  the  bases. 

560.  Find  the  edge  of  a  cube  equivalent  to  a  regular  tetrahedron 
whose  edge  measures  3  inches, 

-  561.  Find  the  edge  of  a  cube  equivalent  to  a  regular  octahedron 
whose  edge  measures  3  inches. 


338  SOLID    GEOMETRY. BOOK   VII. 


Similar  Solids. 

"^  562.  The  dimensions  of  a  trunk  are  4  feet,  3  feet,  2  feet.  What  are  the 
dimensions  of  a  trunk  similar  in  shape  that  will  hold  four  times  as  much  ? 

-  563.  By  what  number  must  the  dimensions  of  a  cylinder  be  multiplied 
in  order  to  obtain  a  similar  cylinder  (i.)  whose  surface  shall  be  n  times 
that  of  the  first;  (ii.)  whose  volume  shall  be  n  times  that  of  the  first? 

564.  A  pyramid  is  cut  by  a  plane  which  passes  midway  between  the 
vertex  and  the  plane  of  the  base.  Compare  the  volumes  of  the  entire 
pyramid  and  the  pyramid  cut  off. 

^  565.  The  height  of  a  regular  hexagonal  pyramid  is  36  feet,  and  one 
side  of  the  base  is  6  feet.  What  are  the  dimensions  of  a  similar  pyramid 
whose  volume  is  -^^  that  of  the  first  ? 

566.  The  length  of  one  of  the  lateral  edges  of  a  pyramid  is  4  meters. 
How  far  from  the  vertex  will  this  edge  be  cut  by  a  plane  parallel  to  the 
base,  which  divides  the  pyramid  into  two  equivalent  parts  ? 

^  567.  The  length  of  a  lateral  edge  of  a  pyramid  is  a.  At  what  dis- 
tances from  the  vertex  will  this  edge  be  cut  by  two  planes  parallel  to 
the  base,  which  divide  the  pyramid  into  three  equivalent  parts  ? 
--  568.  The  length  of  a  lateral  edge  of  a  pyramid  is  a.  At  what  dis- 
tance from  the  vertex  will  this  edge  be  cut  by  a  plane  parallel  to  the 
base,  and  dividing  the  pyramid  into  two  parts  which  are  to  each  other 
as  3:  4? 

569.  The  volumes  of  two  similar  cones  are  54  cubic  feet  and  432  cubic 
feet.     The  height  of  the  first  is  6  feet ;  what  is  the  height  of  the  other  ? 

'  570.  In  each  of  two  right  circular  cylinders  the  diameter  is  equal  to 
the  height.  The  volume  of  one  is  f  that  of  the  other.  What  is  the 
ratio  of  their  heights  ? 

571.  Find  the  dimensions  of  a  right  circular  cylinder  ^f  as  large  as 
a  similar  cylinder  whose  height  is  20  feet,  and  diameter  10  feet. 

'*^572.  The  height  of  a  cone  of  revolution  is  h,  and  the  radius  of  its 
base  is  r.  What  are  the  dimensions  of  a  similar  cone  three  times  as 
large  ? 

^573.  The  height  of  the  frustum  of  a  right  cone  is  |  the  height  of  the 
entire  cone.     Compare  the  volumes  of  the  frustum  and  the  entire  cone. 

'^  574.  The  frustum  of  a  pyramid  is  8  feet  high,  and  two  homologous 
edges  of  its  bases  are  4  feet  and  3  feet  respectively.  Compare  the  vol- 
ume of  the  frustum  and  that  of  the  entire  pyramid.  / 


BOOK  VIII. 

THE     SPHERE. 


Plane  Sections  and  Tangent  Planes. 

679.  A  sphere  is  a  solid  bounded  by  a  surface  every  point 
of  which  is  equally  distant  from  a  point  called  the  centre. 

680.  A  sphere  may  be  generated  by  the  revolution   of  a 
semicircle  ACB  about  its  diameter  AB  as  an  axis. 


681.  A  radius  of  a  sphere  is  a  straight  line  drawn  from  its 
centre  to  its  surface.     < 

682.  A   diameter  of  a   sphere   is   a  straight  line  passing 
through  the  centre  and  limited  by  the  surface. 

Since  all  the  radii  of  a  sphere  are  equal,  and  a  diameter  is 
equal  to  two  radii,  all  the  diameters  of  a  sphere  are  equal. 

683.  A  line  or  plane  is  tangent  to  a  sphere  when  it  has  one, 
and  only  one,  point  in  common  with  the  surface  of  the  sphere. 

684.  Two  spheres  are  tangent  to  each  other  when  their 
surfaces  have  one,  and  only  one,  point  in  common. 


340        /  SOLID   GEOMETRY.  —  BOOK   VIII. 


Proposition  I.     Theorem. 

685.  Every  section  of  a  sphere  made  hy  a  plane  is 
a  circle. 


Let  0  be  the  centre  of  a  sphere,  and  ABD  any  sec- 
tion made  by  a  plane. 

To  prove  that  the  section  ABD  is  a  circle. 

Proof.   Draw  the  radii  OA,  OB,  to  any  two  points  A,  B/\n 
the  boundary  of  the  section,  £lnd  draw  OQ 1.  to  the  section. 
In  the  rt.  A  OAC,  OBC, 

OC  is  common. 

Also  OA  =  OB, 

(being  radii  of  the  sphere), 
.'.AOAO=AOBO,  §161 

.-.  CA  =  CB. 
In  like  manner  any  two  points  in  the  boundary  of  the  sec- 
tion may  be  proved  to  be  equally  distant  from  C. 

Hence  the  section  ABB  is  a  circle  whose  centre  is  C.  .q.  e.  d. 

686i  Cor.  1.  The  line  joining  the  centre  of  a  sphere  to  the 
centre  of  a  circle  of  the  sphere  is_  perpendicular  to  the  plane  of 
the  circle. 

687.  Cor.  2.  Circles  of  a  sphere  made  hy  planes  equally  dis- 
tant from  the  centre  are  equal.  For  AC  =  AO  ~  OC  ;  and 
AO  and  0(7 are  the  same  for  all  equally  distant  circles;  there- 
fore ^C  is  the  same. 


THE   SPHERE.  341 

688.  Cob.  3.  Of  two  circles  made  hy  planes  unequally  distant 
from  the  centre,  the  Tiearer  is  the  larger.  For,  in  the  expression 
AG  =  AO  —00  ,  as  (9 C decreases,  ^Cincreases. 

i9.   A  great  circle  of  a  sphere  is  a  section  made  by  a  plane 
which  passes  through  the  centre  of  the  sphere. 

690.  A  small  circle  of  a  sphere  is  a  section  made  by  a  plane 
which  does  not  pass  through  the  centre  of  the  sphere. 

— ^_691r  The  axis  of  a  circle  of  a  sphere  is  the  diameter  of  the 
sphere  which  is  perpendicular  to  the  plane  of  the  circle.  The 
ends  of  the  axis  are  balled  the  poles. , 

692.  Parallel  circles  have  the  same  axis  and  the  same  poles. 

693.  All  great  circles  of  a  sphere  are  equal. 

694.  Every  greai  circle  bisects  the  sphere.  For  the  two  parts 
into  which  the  sphere  is  divided  can  be  so  placed  that  they 
will  coincide  ;  otherwise  there  would  be  points  on  the  surface 
unequally  distant  from  the  centre. 

695.  Two  great  circles  bisect  each  other.  For  the  intersec- 
tion of  their  planes  passes  through  the  centre,  and  is  a  diameter 
of  each  circle. 

696.  Two  great  circles  whose  planes  are  perpendicular  pass 
through  each  other  s  poles;  and  conversely. 

697.  Through  two  given  points  on  the  surface  of  a  sphere  an 
arc  of  a  great  circle  may  always  be  drawn.  For  the  two  given 
points  together  with  the  centre  of  the  sphere  determine  the 
plane  of  a  great  circle  whose  circumference  passes  through 
the  two  given  points. 

If  the  two  given  points  are  the  ends  of  a  diameter,  the  posi- 
tion of  the  circle  is  not  determined ;  for  through  a  diameter 
an  indefinite  number  of  planes  may  be  passed. 

698i  Through  three  given  points  on  the  surface  of  a  sphere 
one  circle  m,ay  be  drawn,  and  only  one.  For  the  three  points 
determine  one,  and  only  one,  plane.  y 


SOLID    GEOMETRY.  —  BOOK   VIII. 


Proposition  II.     Theorem. 

?99P  The  shortest  distance  on  the  surface  of  a  sphere 
between  any  two  points  on  that  surface  is  the  arc ^  not 
greater  than  a  semi-circumference,  of  the  great  circle 
which  joins  them. 


Let  AB  be  the  arc  of  a  great  circle  which  Joins  any 
two  points  A  and  B  on  the  surface  of  a  sphere;  and 
let  ACPQB  be  anj  other  line  on  the  surface  between 
A  and  B. 

To  prove  ACFQB  >  AB. 

Proof.  Let  P  be  any  point  in  ACFQB. 

Let  arcs  of  great  circles  pass  through  A,  P,  and  P,  B.  §  697 

Join  A,  P,  and  B  with  the  centre  of  the  sphere  0. 
The  A  AOB,  AOP,  and  FOB  are  the  face  A  of  the  trihe- 
dral angle  whose  vertex  is  at  0. 
'  The  arcs  AB,  AF,  and  FB  are  measures  of  these  A.  §  262 
Now  Z  AOF+Z  FOB  is  greater  than  Z  AOB,  §  539 
.-.  arc  AF-\-  arc  FB  >  arc  AB. 
In  like  manner,  joining  any  point  in  ^CP  with  A  and  P, 
and  any  point  in  FQB  with  P  and  P,  by  arcs  of  great  (D,  the 
sum  of  these  arcs  will  be  greater  than  arc  AF-\-  arc  FB ; 
and  therefore  greater  than  arc  AB. 

If  this  process  be  indefinitely  repeated,  the  sum  of  the  arcs 
of  the  great  (D  will  increase  and  always  be  greater  than  AB. 
Therefore  ACFQB,  which  is  the  limit  of  the  sum  of  these 
arcs,  is  greater  than  AB.  q.e.d. 


THE   SPHERE. 


343 


700.   By  the  distance  between  two  points  on  the  surface  of 
a  sphere  is  meant  the  arc  of  a  great  circle  joining  them. 


Proposition  III.     Theorem. 

701.   The  distances  of  all  points  in  the  circumfer- 
ence of  a  circle  of  a  sphere  from  its  poles  are  equal. 


Let  P,  P'  be  the  poles  of  the  circle  ABC,  and  A,  B,  C, 
any  points  on  its  circumference. 

To  prove  that  the  great  circle  arcs  PA,  PB,  PC  are  equal. 

Proof.   The  straight  lines  PA,  PB,  PC  are  equal,        §  478 

Therefore  the  arcs  PA,  PB,  PC  sere  equal.        §  230 

In  like  manner,  the  great  circle  arcs  PA,  PB,  PC  may 
be  proved  equal.  q^  e.  d. 

702.  The  distance  from  the  nearer  pole  of  a  circle  to  any 
point  in  the  circumference  of  the  circle  is  called  the  polar  dis- 
tance of  the  circle. 

703.  Cor.  1.  The  polar  distance  of  a  great  circle  is  a  quad- 
rant-arc. For  it  is  the  measure  of  a  right  angle  whose  vertex 
is  at  the  centre  of  the  sphere. 

704.  Scholium.  The  distances  of  all  points  in  the -circum- 
ference of  a  circle  of  a  sphere  from  any  point  in  its  axis  are 
equal. 


344 


SOLID   GEOMETRY.  —  BOOK   VIII. 


Proposition  IV.     Theorem. 

705.  A  -point  on  the  surface  of  a  sphere,  which  is 
at  the  distance  of  a  quadrant  from  each  of  two  other 
points,  not  the  extremities  of  a  diameter,  is  a  pole  of 
the  great  circle  passing  through  these  points* 


Let  the  distances  PA  and  PB  be  quadrants. 

To  prove  P  a  pole  of  the  great  circle  which  passes  through 
A  and  B. 

Proof.  The  A  POA  and  POP  are  rt.  A, 

{because  each  is  measured  by  an  arc  equal  to  a  quadrant). 

.-.  PO  is  X  to  the  plane  of  the  O  ABO,  §  472 

Hence  P  is  a  pole  of  the  O  ABO.  §  691 

a  E.  D. 

706.  Cor.  The  above  theorem  enables  us  to  describe  with  the 
compasses  an  arc  of  a  great  circle  through  two  given  points 
A  and  B  of  the  surface  of  a  sphere.  For,  if  with  A  and  B  as 
centres,  and  an  opening  of  the  compasses  equal  to  the  chord 
of  a  quadrant  of  a  great  circle,  we  describe  arcs,  these  arcs 
will  cut  at  a  point  P,  which  will  be  the  pole  of  the  great  circle 
passing  through  A  and  B.  Then  with  P  as  centre,  the  arc 
passing  through  A  and  B  may  be  described. 

In  order  to  make  the  opening  of  the  compasses  equal  to  the 
chord  of  a  quadrant  of  a  great  circle,  the  radius  or  the  diam- 
eter of  the  sphere  must  be  given. 


THE   SPHERE. 


345 


Proposition  V.     Problem. 


Given  a  material  sphere  to  find  its  radius. 


:->5 


C> 


V 


Let  PBP'G  represent  a  material  sphere. 
It  is  required  to  find  its  diameter, 

Oonstniction.  From  any  point  P  of  the  given  surface,  with 
any  opening  of  the  compasses,  describe  the  circumference  ABC 
on  the  surface.     Then  the  straight  line  PB  is  known. 

Take  any  three  points  A,  B,  and  C  in  this  circumference, 
and  with  the  compasses  measure  the  chord  distancea^^,  BC^ 
and  CA. 

Construct  the  A  A'B'C\  with  sides  equal  respectively  to 
AB,  BC,  and  CA,  and  circumscribe  a  O  about  the  A  A'B'C. 

The  radius  B'B'  of  this  0  is  equal  to  the  radius  of  O  ^^C. 

Construct  the  rt.  A  bdp,  having  the  hypotenuse  hp  =  BP, 
and  one  side  bd=  B^U. 

Draw  hfp^  X  to  h'p,  and  meeting  pd  produced  in  p'. 

Then  jop'  is  equal  to  the  diameter  of  the  given  sphere. 

Proof.  Suppose  the  diameter  PP  and  the  straight  line  PB 
drawn.  r^^^  ^  ^^p  ^^^  ^^^  ^^^  ^^^^1^  g  ^g^ 

Hence  the  A  PBP  and  pip'  are  equal.  §  149 

Therefore />/  =  PP'. 
And  \pp^  is  equal  to  the  radius.  q.  e.f. 


346 


SOLID   GEOMETRY.  —  BOOK    VIII. 


Proposition  VI.     Theorem. 

708.  A  plane  perpendicular  to  a  radius  at  its  ex- 
tremity is  tangent  to  the  sphere. 


Let  0  be  the  centre  of  a  sphere,  and  MN  a  plane 
perpendicular  to  the  radius  OP,  at  its  extremity  P 

To  prove  MN  tangent  to  the  sphere. 

Proof.    From  0  draw  any  other  straight  line   OA  to  the 

plane  MN. 

OF<OA,  §477 

(a  A.  is  the  shortest  distance  from  a  point  to  a  plane). 

Therefore  the  point  A  is  without  the  sphere. 
Similarly  we  may  prove  that  every  point,  except  P,  in  the 
plane  MNis  without  the  sphere, 

Therefore  MN  is  tangent  to  the  sphere  at  P.      §  683 

a  E.  D. 

709.  Cor.  1.  A  plane  tangent  to  a  sphere  is  perpendicular 
to  the  radius  drawn  to  the  point  of  contact. 

710.  Cor.  2.  A  straight  line  tangent  to  a  circle  of  a  sphere 
lies  in  a  plane  tangent  to  the  sphere  at  the  point  of  contact.^  4.7^ 

711.  Cor.  3.  Any  straight  line  in  a  tangent  plane  through 
the  point  of  contact  is  tangent  to  the  sphere  at  that  point. 

712.  Cor.  4.    The  plane  of  tivo  straight  lines  tangent  to  a 
sphere  at  the  sanne  point  is  tangent  to  the  sphere  at  that  point. 


THE   SPHERE.  347 

713.  A  sphere  is  said  to  be  inscribed  in  a  polyhedron  when 
all  the  faces  of  the  polyhedron  are  tangent  to  the  sphere. 

714.  A  sphere  is  said  to  be  circumscrihed  about  a  polyhe- 
dron when  all  the  vertices  of  the  polyhedron  lie  in  the  surface 
of  the  sphere. 

Proposition  VII.     Theorem. 

715.  A  sphere  may  be  inscribed  in  any  given  tetror- 
hedron,  D 


B 

Let  ABCD  he  the  given  tetrahedron. 

To  prove  that  a  sphei-e  may  be  inscribed  in  ABOD. 

Proof.  Bisect  the  dihedral  A  at  the  edges  AB,  BC,  and  AC 
by  the  planes  OAB,  OBC,  and  OAC,  respectively. 

Every  point  in  the  plane  OAB  is  equally  distant  from  the 
faces  ABC  and  ABD.  -  "^  §  525 

For  a  like  reason,  every  point  in  the  plane  OBC  is  equally 
distant  from  the  faces  ^^Cand  DBC\  and  every  point  in  the 
plane  OAC\?>  equally  distant  from  the  faces  ^jSCand  ADC. 

Therefore  0,  the  common  intersection  of  these  three  planes, 
is  equally  distant  from  the  four  faces  of  the  tetrahedron. ^=**' 

Hence  a  sphere  described  with  0  as  a  centre,  and  with  the 
radius  equal  to  the  distance  from  0  to  any  face,  will  be  tangent 
to  each  face,  and  will  be  inscribed  in  the  tetrahedron.      §  713 

a&o. 

716.  Cor.  The  six  planes  which  bisect  the  six  dihedral  angles 
of  a  tetrahedron  intersect  in  the  same  point. 


348 


SOLID    GEOMETRY.  —  BOOK    VIII. 


Proposition  VIII.     Theorem. 

717.  A    sphere    may    he  circumscribed    about  any 
given  tetrahedron. 


Let  ABCD  be  the  given  tetrahedron. 

To  prove  that  a  sphere  may  he  circumscrihed  about  ABCD. 
Proof.   Let  M^  N,  respectively  be  the  centres  of  the  circles 
circumscribed  about  the  faces  ABC,  ACD. 

Let  also  MR  be  ±  to  face  ABC,  NS 1.  to  face  ACD. 
MR  is  the  locus  of  points  equidistant  from  A,  B,  C, 
and  N8  is  the  locus  of  points  equidistant  from  A,  C,  D.^  480 

Also  MR  and  JVS  lie  in  the  same  plane. 

For,  if  a  plane  -L  to  AC  he  passed  through  its 'middle 

point,  this  plane  will  contain  all  points  equidistant  from  A 

and  O.  §  482 

.*.  MR  and  I^S  must  lie  in  this  plane. 

Also  MR  and  iV/S',  being  X  to  planes  which  are  not  II,  can- 
not be  II,  and  must  therefore  meet  at  some  point  0. 
.'.  O  is  equidistant  from  A,  B,  C,  and  D, 
and  a  spherical  surface  whose  centre  is  0,  and  radius  OA. 
will  pass  through  the  points  A,  B,  C,  and  D.  a  e.  o 

718.  Cor.  1.   The  four  perpendiculars  erected  at  the  centres 
of  the  faces  of  a  tetrahedron  meet  at  the  same  point. 

719.  Cor.  2.   The  six  planes  perpendicular  to  the  edges  of  a 
tetrahedron  at  their  middle  points  intersect  at  the  same  point. 


THE   SPHERE.  349 


Proposition  IX.     Theorem. 

720.  The  intersection  of  two  spherical  surfaces  is 
the  circumference  of  a  circle  whose  plane  is  perpen- 
dicular to  the  line  joining  the  centres  of  the  surfaces 
and  whose  centre  is  in  that  line. 


Let  0,  0'  be  the  centres  at  the  spherical  surfaces, 
and  let  a  plane  passing  through  0,  0'  cut  the  sphere 
in  great  circles  whose  circumferences  intersect  each 
other  in  the  points  A  and  B. 

To  prove  that  the  spherical  surfaces  intersect  in  the  circ7im- 
ference  of  a  circle  whose  plane  is  perpendicular  to  00',  and 
whose  centrt  is  the  point  0  where  AB  meets  OO. 

Proof.  The  common  chord  AB  is  X  to  00'  and  bisected 
at  (7,  §  249 

{when  two  circumferences  intersect  each  other,  the  line  joining  their  centres 
is  J-  to  the  common  chord  at  its  middle  point). 

If  the  plane  of  the  two  great  circles  revolve  about  00',  their 
circumferences  will  generate  the  two  spherical  surfaces,  and 
the  point  A  will  describe  the  line  of  intersection  of  the  surfaces. 

But  during  the  revolution  ^C  will  remain  constant  in  length 
and  ±  to  00'. 

Therefore  the  line  of  intersection  described  by  the  point  A 
will  be  the  circumference  of  a  circle  whose  centre  is  C  and 
whose  plane  is  ±  to  00',  §  473 

Q.E.D. 


350 


SOLID    GEOMETRY. BOOK    VIII. 


Figures  on  the  Surface  of  a  Sphere. 

721.  The  angle  of  two  curves  passipg  through  the  same  point 
is  the  angle  formed  by  the  two  straight  lines  tangent  to  the 
curves  at  that  point.  If  the  two  curves  are  arcs  of  great  cir- 
cles, th^  angle  is  called  a  s'pherical  angle. 

Proposition  X.     Theorem. 

(723/  A  spherical  angle  is  measured  hy  the  arc  of 
a  great  circle  described  from  its  verteoc  as  a  pole  and 
included  between  its  sides  (produced  if  necessary). 


Let  ABy  AC  be  arcs  of  great  circles  intersecting  at 
A;  AB'  and  AC,  the  tangents  to  these  arcs  at  A;  EC 
an  arc  of  a  great  circle  described  from  A  as  a  pole 
and  included  between  AB  and  AC. 

To  prove  that  the  spherical  A  BACis  measured  hy  arc  BC. 
Proof.  Draw  the  radii  OA,  OB,  00. 

In  the  plane  AOB,  AB'  is  ±  to  AO, 


and 


§240 
§100 


OB  is  ±  to  AO. 
.:  AB'  is  II  to  OB. 
Ada  II  to  00. 
.'.Z  B'AC'=A  BOO. 
Z  BOO    is  measured  by  arc  BO. 
.'.  Z  B'AO'  is  measured  by  arc  BO. 
.'.  Z  BAO    is  measured  by  arc  ^(7.  q. e. o. 

723.    Cor.  A  spherical  angle  has  the  same  measure  as  the 
dihedral  angle  formed  hy  the  planes  of  the  two  circles. 


Similarly, 


But 


§498 
§262 


THE   SPHERE.  351 


Proposition  XL     Problem. 

724.  To  describe  an  arc  of  a  great  circle  through  a 
given  point  perpendicular  to  a  given  arc  of  a  great 
circle. 


Let  A  be  a  point  on  the  surface  of  a  sphere,  CHD 
an  arc  of  a  great  circle,  P  its  pole. 

To  describe  an  arc  of  a  great  circle  through  A  perpendicular 
to  CHD. 

Oonstniction.  From  ^  as  a  pole  describe  an  arc  of  a  great 
circle  cutting  CHD  at  E. 

From  ^  as  a  pole  describe  the  arc  AB  through  A. 

Then  ^^  is  the  arc  required. 

Proof.  The  arc  ^^  is  the  arc  of  a  great  circle,  and  E  is  its 
pole  by  construction.  §  703 

The  point  E  is  at  the  distance  of  a  quadrant  from  P.  §  703 

Therefore  the  arc  AB  produced  will  pass  through  P. 

And  since  the  spherical  Z  PBE  is  measured  by  an  arc  of  a 
great  circle  extending  from  P  to  E,  §  722 

the  Z  ABD  is  a  right  angle. 

Therefore  the  arc  AB  is  J-  to  the  arc  CHD.       q  e  d 


Ex.  575.  Every  point  in  a  great  circle  which  bisects  a  given  arc  of  a 
great  circle  at  right  angles,  is  equidistant  from  the  extremities  of  the 
given  arc. 


352  SOLID   GEOMETRY. — BOOK   VIII. 

"725.  A  spherical  polygon  is  a  portion  of  the  surface  of  a 
sphere  bounded  by  three  or  more  arcs  of  great  circles. 

The  bounding  arcs  are  the  ddes  of  the  polygon  ;  the  angles 
which  they  form  are  the  angles  of  the  polygon ;  their  points^ 
of  intersection  are  the  vertices  of  the  polygon. 

The  values  of  the  sides  of  a  spherical  polygon  are  usually 
expressed  in  degrees,  minutes,  and  seconds. 

726.  The  planes  of  the  sides  of  a  spherical  polygon  form  a 
polyhedral  angle  whose  vertex  is  the  centre  of  the  sphere, 
whose  face  angles  are  measured  by  the  sides  of  the  polygon, 
and  whose  dihedral  angles  have  the  same  numerical  measure 
as  the  angles  of  the  polygon. 

Thus,  the  planes  of  the  sides  of  the  polygon  ABCD  form 
the  polyhedral  angle  0-ABCD.  The  face 
angles  AOB,  BOC,  etc.,  are  measured  by 
the  sides  AB,  BC,  etc.,  of  the  polygon. 
The  dihedral  angle  whose  edge  is  OA 
has  the  same  measure  as  the  spherical 
angle  BAD,  etc. 

Hence, /rom  any  property  of  polyhedral 
angles  we  may  infer  an  analogous  property  of  spherical  poly- 
gons; and  conversely. 

727.  A  spherical  polygon  is  convex  if  the  corresponding 
polyhedral  angle  is  convex  (§  534).  Every  spherical  polygon 
is  to  be  assumed  convex  unless  otherwise  stated. 

728.  A  diagonal  of  a  spherical  polygon  is  an  arc  of  a  great 
circle  connecting  any  two  vertices  which  are  not  adjacent. 

729.  A  spherical  triangle  is  a  spherical  polygon  of  three 
sides ;  like  a  plane  triangle,  it  may  be  right  or  oblique,  equi- 
lateral, isosceles,  or  scalene. 

730.  Two  spherical  polygons  are  equal  if  they  can  be  applied, 
the  one  to  the  other,  so  as  to  coincide. 


THE   SPHERE. 


353 


Proposition  XII.     Theorem. 

731   Each  side  of  a  spherical  triangle  is  less  than 
the  sum  of  the  other  two  sides. 


LetABG  be  a  spherical  triangle,  AB  the  largest  side. 

To  prove  AJB  <  A0+  BO, 

Proof.   In  the  corresponding  trihedral  angle  0-ABC, 

^  AOB  is  less  than  Z  AOC+  Z.  BOO.  §  539 

.•.AB<AO-\-BC.  §726 

Q.B.a 

Proposition  XIII.    Theorem. 

732.  The  sum  of  the  sides  of  a  spherical  polygon  is 
less  than  S6(P, 


Let  ABCD  be  a  spherical  polygost 

To  prove       AB  +  B0+  CD -\- DA  <  360°. 
Proof.    In  the  corresponding  polyhedral  angle  O-ABCB,  the 
sum  of  all  the  face  angles  is  less  than  360°.  §  540 

.-.  AB+  B0-\-  CD^DA<  360". 


aE.a 


,  /  354  SOLID    GEOMETRY.  —  BOOK    VIII. 

733.  If,  from  the  vertices  of  a  spherical  triangle  as  poles, 
\)      arcs  of  great  circles  are  described,  a   spherical   triangle   is 

formed,  which  is  called  the  jpolar  triangle  X 

of  the  first.  Thus,  if  ^,  ^,  Q  are  the  poles 
of  the  arcs  of  the  great  circles  B^C\  A'C\ 
A^B\  respectively,  then  A^BW^  is  the 
polar  triangle  of  ABC. 

If,  with  A^  B^  C  &s  poles,  entire  great    -^  ^ (^ 

circles  instead  of  arcs  are  described,  these  circles  will  divide 
the  surface  of  the  sphere  into  eic/ht  spherical  triangles. 

Of  these  eight  triangles,  that  one  is  the  polar  of  J.5C  whose 
vertex  A^,  corresponding  to  A,  lies  on  the  same  side  of  BO  asy 
the  vertex  A  ;  and  similarly  with  the  other  vertices. 

Proposition  XIV.     Theorem. 

734.  If  A'B'C  is  the  polar  triangle  of  ABC,  then, 
reciprocally,  ABC  is  the  polar  triangle  of  A^B'C\ 

A' 


'0 
Let  AfB'C  be  the  polar  triangle  of  ABC. 

To  prove  that  ABC  is  the  polar  triangle  of  A^B^C. 
Proof.  Since  A  is  the  pole  of  B'C\  §  733 

.*.  B'  is  at  a  quadrant's  distance  from  A.  §  703 

Similarly,  since  C  is  the  pole  of  A^B\ 
,'.  B*  is  at  a  quadrant's  distance  from  C. 

.'.  B^  is  the  pole  of  the  arc  AC.  §  705 

Similarly,  A*  is  the  pole  of  BC,  and  C^  the  pole  of  AB. 

-   .'.  ABCis  the  polar  triangle  of  A'B'C  §  733 


aE.  D. 


THE   SPHERE.  355 


Proposition  XV.     Theorem. 

In,  two  polar  triangles  eaxih  angle  of  the  one 
is  the  supplement  of  the  opposite  side  in  the  other. 


LetABCyA'B'Obe  two  polar  triangles;  then  let  the 
letter  at  the  vertex  of  each  angle  denote  its  value 
in  angle- degrees,  and  the  corresponding  small  letters 
the  values  of  the  opposite  sides  in  arc- degrees. 

To  prove  A  +a'  =  180°,    B  -{-b'=  180°,    C  +c'  =  180°. 

A'  +  a=  180°,    B'-i-b  =  180°,    C*-\-c  =  180°. 

Proof.  Produce  the  arcs  AB,  AC  until  they  meet  B'C  at 
the  points  D,  E,  respectively. 

Since  B^  is  the  pole  of  AE,  B^E=  90°. 
Since  (7'  is  the  pole  of  AB,  C"Z>-=  90°. 
Adding,  we  have    B'E+  C*B  =  180°. 
That  is,  B'l)  +DE-\^C'n  =  180°. 

Or  Z)^+5'e'  =  180°. 

'EutBW^  =  a!. 
Also  DE  measures  /.A,  §  722 

.-.  A-\-a'  =  180°. 
In  a  similar  way  all  the  other  relations  are  proved.       q,  e.  d. 

736.  Scholium.  Two  polar  triangles  are  sometimes  called 
supplemental  triangles. 


356  SOLID    GEOMETRY.  —  BOOK    VIII. 


Proposition  XVI.     Theorem. 

7§7^  The  sum  of  the  angles  of  a  spherical  triangle 
is  greater  than  18(P  and  less  than  540^. 


a' 
Let  ABC  be  a  spherical  triangle,  and  let  A,  B,  C  denote 
the  values  of  its  angles,  and  a',  b',  c',  respectively,  the 
values  of  the  opposite  sides  in  the  polar  triangle  A'B'CK 

To  prove       A-\-B-\-  C>  180^  and  <  540°. 
Proof.    Since  the  A  ABC,  A^B^C\  are  polar  A, 

A^a^=  180°,  B-i-b'=  180°,  C+  c'  =  180°.     §  735 
By  addition,  A-i- B-\-0+a'+b^ +  c' =  540°. 

.-.  ^  +  ^+(7-540°-(a'+^»'  +  c'). 
Now  a'  +  5'  +  c'  is  less  than  360°,  §  732 

.\A-\-B-{-0=  540°  -  some  number  less  than  360°. 
.\A  +  B+C>  180°. 
And  since  a'  +  Z>'  -j-  <?'  is  greater  than  0°, 

.;A  +  B+0<5W,    .  ■     Q.E.D. 

738.  OoR.  A  spherical  triangle  may  have  two,  or  even  three^ 
right  angles  ;  and  it  may  have  two,  or  even  three,  obtuse  angles. 

739.  A  spherical  triangle  having  two  right  angles  is  called 
a  hi-rectangular  triangle;  and  a  spherical  triangle  having 
three  right  angles  is  called  a  tri-rectangular  triangle. 

740.  The  difference  between  the  sum  of  the  angles  of  a 
spherical  triangle  and  180°  is  called  the  spherical  excess  of 
the  triangle. 


THE  SPHERE.  357 


Proposition  XVII.     Theorem. 

741.  In  a  hi-rectangular  spherical  triangle  the  sides 
opposite  the  right  angles  are  quadrants,  and  the  side 
opposite  the  third  angle  measures  that  angle. 

A 


Let  ABC  be  a  bi-rectangnlar  spherical  triangle,  with 
the  angles  at  B  and  C  right  angles. 

To  prove  that  AB  and  AC  are  qiLadrants^  and  that  Z  A  is 
measured  by  BC. 

Proof.  Since  the  A  B  and  C  are  right  angles,  the  planes  of 
the  arcs  AB,  AC qxq  X  to  the  plane  of  the  arc  BC.         §  723 

.'.  AB  and  ^(7  must  each  pass  through  the  pole  of  BO,  §  696 
{two  great  circles  whose  planes  are  ±  pass  through  each  other's  poles). 

.'.  A  is  the  pole  of  BC. 

.'.  AB  and  AC  are  quadrants,  §  703 

and  Z  ^  is  measured  by  the  arc  BC  §  722 

aE.  D. 

742.  Cor.  1.  If  two  sides  of  a  spherical  triangle  are  quad- 
rants, the  third  side  measures  the  opposite  angle. 

743.  Cor.  2.  Each  side  of  a  tri-rectan^ular  spherical  triangle 
is  a  quadrant. 

744.  Cor.  3.  Three  planes  passed  through  the 
centre  of  a  sphere,  each  perpendicular  to  the  other 
two  planes,  divide  the  surface  of  the  sphere  into 
eight  tri-rectangular  triangles. 


358 


SOLID    GEOMETEY. BOOK    VIII. 


745.  If  through  the  centre  0  of  a  sphere  three  diameters 
A  A',  BB\  CC^  are  drawn,  and  the  points  A,  B,  C  sue  joined 
by  arcs  of  great  circles,  and  also  the 
points  A\  B\  C\  the  two  spherical 
triangles  ^^(7  and  A^B^C*  are  called 
symmetrical  spherical  triangles. 

The  corresponding  trihedral  angles 
are  also  symmetrical.  §  538 

In  the  same  way  we  may  form  two 
symmetrical  polygons  of  any  number 
of  sides.     And  after  they  are  formed  they  may  be  placed  in 
any  positions  upon  the  surface  of  the  sphere. 

746.  Two  symmetrical  triangles  are  mutually  equilateral 
and  equiangular ;  yet  in  general  they  cannot  be  made  to  coin- 
cide by  superposition.  If  in  the  above 
figure,  the  hemisphere  below  the  great 
circle  BCB^Q^  be  revolved  about  its 
axis  through  half  a  revolution,  the  tri- 
angle A^B^O*  will  take  the  position 
A'^BC,  and  it  will  now  be  quite  evident  that  the  triangles 
cannot  be  made  to  coincide.  If  the  triangles  are  placed  so  that 
B^O*  coincides  with  CB,  and  A  and  A'  lie  on  the  same  side 
of  BO,  it  will  be  seen  that  the  equal  parts  of  the  two  triangles 
occur  in  reverse  order. 

747.  If,  however,  AB  =  AC,  and  A' B^  =  A^ C^ ;  th&t  is,  if 
the  two  symmetrical  triangles  are 
isosceles,  then,  because  AB,  AC, 
A^B\  A^C*,  are  all  equal,  and  the 
angles  A  and  A^  are  equal,  being 
opposite  dihedral  angles  (§  745),  the 
two  triangles  can  be  made  to  coin- 
cide ;  in  other  words. 

If  two  symmetrical  spherical  triangles  are  isosceles,  they  are 
superposable,  and  therefore  equal. 


THE    SPHERE.  359 


Proposition  XVIII.     Theorem. 

748.  Two  symmetrical  spherical  triangles  are  equiv- 
alent. 


Let  ABC,  A'S^C  be  two  symmetrical  spherical  tri- 
angles  with  their  homologous  vertices  diametrically 
opposite  to  each  other. 

To  prove  that  the  triangles  ABC,  A'B'C  are  equivalent. 

Proof.  Let  JP  be  the  pole  of  a  small  circle  passing  through 
the  points  A,  B,  C,  and  let  FOB'  be  a  diameter. 

Draw  the  great  circle  arcs  FA,  FB,  FC,  FA',  FB',  FC\ 
FA=  FB=  FC.  §  701 

And  since      F'A' =  FA,   FB' =  FB,   F'C'=FC,   §746 
.-.  F'A'  =  FB'=FC'. 
The  two  symmetrical  A  FAC,  FA'C  are  isosceles. 

.\AFAC=AFA'C'.  §747 

Similarly,  A  FAB  =  A  F'A'B', 

and  AFBC=AFB'C'. 

Now       A  ABC=^  A  FAC-}- A  FAB  +  A  FBC, 
and       A  A'B'C  =c=  A  FA'C  +  A  FA'B'  +  A  FB'C. 

'.AABC^AA'B'C.  aE.D. 

If  the  pole  F  should  fall  without  the  A  ABC,  then  F  would 

fall  without  A  A'B'C,  and  each  triangle  would  be  equivalent 

to  the  sum  of  t^o  isosceles  triangles  diminished  by  the  third ; 

so  that  the  result  would  be  the  same  as  before.  / 


P60  SOLID   GEOMETRY. BOOK    VIII. 


a 


Proposition  XIX.     Theorem. 


749i  Two  triangles  on  the  same  sphere,  or  equal 
spheres,  are  equal  or  equivalent,  if  two  sides  and  the 
included  angle  of  the  one  are  respectively  equal  to 
two  sides  and  the  included  angle  of  the  other. 


I.  In  the  triangles  ABC  and  DEF  let  angle  A  equal 
angle  D,  and  the  sides  AB  and  AC  equal  respectively 
the  sides  DE  and  DF;  and  let  the  parts  of  the  two 
triangles  be  arranged  in  the  same  order. 

To  prove  triangles  ABC  and  DEF  equal. 
Proof.   A  ABQ  can  be  applied  to  A  DEF,  as  in  the  corre- 
sponding case  of  plane  A,  and  will  coincide  with  it.         §  150 

II.  In  the  triangles  ABO  and  D'E'F'  let  angle  A  equal 
angle  D',  and  the  sides  AB  and  AG  equal  respectively 
the  sides  D'E'  and  D'P ;  and  let  the  parts  of  the  two 
triangles  "be  arranged  in  reverse  order. 

To  prove  triangles  ABQ  and  D^EF^  equivalent. 
Proof,   Let  the  A  DEF  upon  the  same  or  an  equal  sphere 
be  symmetrical  with  respect  to  the  A  D^EFK 

Then  A  DEF  has  its  A  and  sides  equal  respectively  to 
those  of  the  A  D'E^FK 

Also  in  the  A  ^^(7  and  DEF 

AA  =  AD,  AB  =  DE,  AQ^DF, 
and  the  parts  are  arranged  in  the  same  order. 

:.t.ABC  =  /^  DEF.  Case  I. 

But  A  DE'F'^  A  DEF,  §  748 

.-.AABO^AD'E'F^,  ^^o. 


'.aU 


THE   SPHERE.  361 


Proposition  XX.     Theorem. 


750.  Two  triangles  on  the  same  sphere,  or  equal 
spheres,  are  equal  or  equivalent,  if  a  side  and  two 
adjacent  angles  of  the  one  are  equal  respectively  to  a 
side  and  two  adjacent  angles  of  the  other. 


Proof.  One  of  the  A  may  be  applied  to  the  other,  or  to  its 
symmetrical  A,  as  in  the  corresponding  case  of  plane  A.  §  147 

Proposition  XXI.     Theorem, 

751.  Two  mutually  equilateral  triangles  on  the 
same  sphere,  or  equal  spheres,  are  mutually  equian- 
gular, and  are  equal  or  equivalent. 

Proof.  The  face  A  of  the  corresponding  trihedral  A  at  the 
centre  of  the  sphere  are  equal  respectively, 

{dnce  they  are  measured  hy  equal  sides  of  the  ^. 

Therefore  the  corresponding  dihedral  A  are  equal.  §  542 
Hence  the  A  of  the  spherical  A  are  respectively  equal. 
Therefore  the  A  are  either  equal,  or  symmetrical  and  equiv- 
alent, according  as  their  equal  sides  are  arranged  in  the  same 
or  reverse  order.  a  e.  ot 

Ex.  576.  The  radius  of  a  sphere  is  4  inches.  From  any  point  on  the 
surface  as  a  pole  a  circle  is  described  upon  the  sphere  with  an  opening 
of  the  compasses  equal  to  3  inches.    Find  the  area  of  this  circle,  y' 


362  SOLID    GEOMETRY. BOOK    VIII. 


Proposition  XXII.     Theorem. 

752,  Two  mutually  equiangular  triangles,  on  the 
same  sphere,  or  equal  spheres,  are  mutually  equilat- 
eral, and  are  either  equal  or  equivalent. 


Let  the  spherical  triangles  T  and  T'  he  mutually 
equiangular. 

To  prove  triangles  T  and  V  mutually  equilateral,  and  equal 
or  equivalent. 

Proof.  Let  A  P  and  P  be  the  polar  A  of  the  A  T  and  T\ 
respectively. 

The  A  P  and  P'  are  mutually  equilateral,  because  in  two 
polar  A  each  side  of  the  one  is  the  supplement  of  the  angle 
lying  opposite  to  it  in  the  other.  §  735 

.'.  A  P  and  P'  are  mutually  equiangular,  because  two 
mutually  equilateral  A  on  equal  spheres  are  mutually  equi- 
angular. §  761 
.'.A  J' and  T'  are  mutually  equilateral. 

Hence  A  T  and  T^  are  either  equal,  or  symmetrical  and 
equivalent,  because  two  mutually  equilateral  A  on  equal 
spheres  are  either  equal,  or  symmetrical  and  equivalent.  §  751 

Q.  E.  D. 


Remark.  The  statement  that  mutually  equiangular  spherical  triangles 
are  mutually  equilateral,  and  equal,  or  equivalent,  is  true  only  when 
limited  to  the  same  sphere,  or  equal  spheres.  But  when  the  spheres  are 
unequal,  the  spherical  triangles  are  unequal ;  and  the  ratio  of  their 
homologous  sides  is  equal  to  the  ratio  of  the  radii  of  the  spheres  on 
which  they  are  situated.   (^  427.) 


THE    SPHERE.  363 


Proposition  XXIII.     Theorem. 

(  78^.  In  an  isosceles  spherical  triangle,  the  angle 
opposite  the  equal  sides  are  equal* 


D 

In  the  spherical  triangle  ABC^  let  AB. equal  AC. 

To  prove  A  B  = /.  C. 

Proof.  Draw  arc  AD  of  a  great  circle,  from  the  vertex  A 
to  the  middle  of  the  base  BC. 

Then  A  ABD  and  ACD  are  mutually  equilateral. 

.*.  A  ABD  and  ACD  are  mutually  equiangular,  §  751 
{two  mutually  equilateral  A  on  the  same  sphere  are  mutually  equiangular). 

.■.ZB  =  ZC, 

{since  they  are  homologous  A  of  symmetrical  A). 

Q.  E.  D. 

754.  Cor.  The  arc  of  a  great  circle  drawn  from  the  vertex 
of  an  isosceles  spherical  triangle  to  the  middle  of  the  base  bisects 
the  vertical  angle,  is  perpendicular  to  the  base,  and  divides  the 
triangle  into  two  symmetrical  triangles. 


Ex.  577.   At  a  given  point  in  a  given  arc  of  a  great  circle,  to  con- 
struct a  spherical  angle  equal  to  a  given  spherical  angle. 

Ex.  578.   To  inscribe  a  circle  in  a  given  spherical  triangle. 

'   Ex.  579.   To  circumscribe  a  circle  about  a  given  spherical  triangle. 


364  SOLID  GEOMETRY.  —  BOOK   VIII. 


Proposition  XXIV.    Theorem. 

/^ 

^5y  If  two  angles  of  a  spherical  triangle  are  equal, 

the  sides  opposite  these  angles  are  equal,  and  the 

triangle  is  isosceles. 


In  the  spherical  triangle  ABC,  let  angle  B  equal 
angle  C. 

To  prove  AC=AB. 

Proof.     Let  the  A  A'B'O'  be  the  polar  A  of  the  A  ABC. 
By  hypothesis  Z  B  =  Z  (7, 

.'.A'C^  =  A^B\  §735 

one  is  the  supple'i 
to  it  in  the  other). 

.\ZB'  =  ZC'.  §753 


in  two  polar  A,  each  side  of  one  is  the  supplement  of  the  Z  lying  opposite 

ther). 


'.AC=AB.  §735 

aE.0. 


Ex.  580.  Given  a  spherical  triangle  whose  sides  are  60°,  80°,  and  100°; 
find  the  angles  of  its  polar  triangle. 

Ex.  581.  Given  a  spherical  triangle  whose  angles  are  70°,  75°,  and 
95° ;  find  the  sides  of  its  polar  triangle.  /'     ' 

Ex.  582.  Given  two  mutually  equiangular  triangles  on  spheres  whose 
radii  are  12  inches  and  20  inches  respectively;  find  the  ratio  of  two 
homologous  sides  of  these  triangles.    (See  note,  page  362.) 


THE   SPHERE.  365 


/^  Proposition  XXV.     Theorem. 

756.  If  two  angles  of  a  spJierical  triangle  are  un/- 
equal,  the  sides  opposite  are  unequal,  and  the  greater 
side  is  opposite  the  greater  angle ;  conversely,  if  two 
sides  are  unequal,  the  angles  opposite  are  unequal, 
and  the  greater  angle  is  opposite  the  greater  side. 


L  In  the  triangle  ABC,  let  the  angle  ABC  he  greater 
than  the  angle  ACB, 
To  prove  AC'>  AB. 

Proof.    Draw  the  arc  £D  of  a  great  circle,  making  Z  OJBD 
equal  Z  ACB. 
Then  I>C=  BB,  §  755 

Now  AB-{-DB>AB,  §  731 

,\AI)  +  I)C>  AB,  or  AC  >  AB. 

n.  Let  AC  be  greater  than  AB. 

To  prove  A  ABQ greater  than  Z.  A  CB. 

Proof.   The  Z  ABC  must  be  equal  to,  less  than,  or  greater 
than  the  Z  ACB. 

If  /.  ABC=  Z  C,  then  AC=  AB,  §  755 

and  if       Z  ABCk  less  than  Z  C,  then  A0<  AB.      Case  I. 

But  both  of  these  conclusions  are  contrary  to  the  hypothesis. 

/.  Z  ABOb  greater  than  Z  C.  q,e.b. 


366 


SOLID    GEOMETRY. 


BOOK    VIII. 


Measurement  of  Spherical  Surfaces. 

757.  A  zone  is  a  portion  of  the  surface  of  a  sphere  included 
between  two  parallel  planes. 

The  circumferences  of  the  sections  made  by  the  planes  are 
called  the  bases  of  the  zone,  and  the  distance  between  the 
planes  is  its  altitude. 

758.  A  zone  of  one  base  is  a  zone  one  of  whose  bounding 
planes  is  tangent  to  the  sphere. 

If  a  circle  (Fig.  1)  be  revolved  about  a  diameter  PQ,  the 
arc  AD  will  generate  a  zone,  the  points  A  and  D  will  gen- 
erate its  bases,  and  (91P  is  its  altitude.  The  arc  I* A  will 
generate  a  zone  of  one  base. 

A 


759.  A  lune  is  a  portion  of  the  surface  of  a  sphere  bounded 
by  two  semi-circumferences  of  great  circles. 

760.  The  angle  of  a  lune  is  the  angle  between  the  semi- 
circumferences  which  form  its  boundaries.  Thus  (Fig.  2), 
ABEQA  is  a  lune,  BA  0  is  its  angle. 

761.  As  in  Plane  Geometry  it  is  convenient  to  divide  a 
quadrant  of  a  circle  into  90  equal  parts,  called  degrees,  so  in 
Solid  Geometry  it  is  convenient  to  divide  each  of  the  eight 
equal  tri-rectangular  triangles  of  which  the  surface  of  a  sphere 
is  composed  (§  744)  into  90  equal  parts,  and  to  call  these 
parts  spherical  degrees.  The  surface  of  every  sphere  therefore 
contains  720  spherical  degrees. 


THE    SPHERE. 


367 


Proposition  XXVI.     Theorem. 

762.  The  area  of  the  surface  generated  hy  a  straight 
line  revolving  about  an  axis  in  its  plane  is  equal  to 
the  product  of  the  projection  of  the  line  on  the  axis 
hy  the  circumference  whose  radius  is  a  perpen- 
dicular erected  at  the  middle  point  of  the  line  and 
terminated  hy  the  a^xis. 


J. 

4           M 

B                                  B 

j                     M^^^\ 

1 

1 

\  Y^^^ — ^^ 

! 

i      i          i,\       I 

X  c 

7            0 

DC            0    R      D 

R   1) 


*  Let  XY  be  the  axis,  AB  the  revolving  line,  CD  its 
projection  on  XY,  M  its  middle  point,  MO  perpendicu- 
lar to  XY,  and  MB  perpendicular  to  AB. 

To  prove  that      area  AB  =  CD  X  ^ttMM. 

Proof.  (1)  If  AB  is  11  to  XY,  then  CD  =  AB,  MR  coincidea 
with  MO,  area  AB  is  the  surface  of  a  right  cylinder,  and  the 
truth  of  the  theorem  follows  at  once  from  §  646. 

(2)  If  AB  is  not  )l  to  XY,  area  AB  will  be  the  surface  of 
the  frustum  of  a  cone  of  revolution. 

.-.  area  AB  =  ABx  2irM0,  §  676 

Draw  AE  II  to  XY. 
The  A  ABE,  MOR  are  similar.  §  327 

.-.  MO:  AE^MR:AB. 
.'.  ABxMO=  AExMR. 
Or,  since  AE=  CD,  §  180 

ABxMO=CDxMR. 
Substituting  this  value  of  AB  X  MO  in  the  first  equation, 
we  obtain  area  AB  =  CD  X  2TrMR. 

(3)  If  ^  lies  in  the  axis  XY,  the  above  reasoning  still  holds 
good  ;  only  AE  and  CD  coincide,  and  the  truth  follows  from 
§670.  aE.a 


^ 


368  SOLID  GEOMETRY.  —  BOOK   VIIL 


Proposition  XXVII.    Theorem. 

^7§^  The  area  of  the  surface  of  a  sphere  is  equal  to 
the  product  of  its  diameter  by  the  circumference  of 
a  great  circle. 


Let  the  sphere  be  generated  by  the  semicircle 
ABODE  revolving:  about  the  diameter  AE,  and  let  0  be 
the  centre,  and  B  the  radius. 

To  prove  that  the  area  of  the  surface  =  AU  X  27ri?. 

Proof.  Inscribe  in  the  semicircle  half  of  a  regular  polygon 
having  an  even  number  of  sides,  as  ABODE. 

From  the  centre  draw  Js  to  the  chords  AB,  BC,  etc. 
These  Js  bisect  the  chords  (§  232)  and  are  equal  (§  236). 
Let  a  denote  the  length  of  each  of  these  ^, 
From  B  and  D  drop  the  Js  ^i^and  DO  to  AE. 
When  the  semicircle  revolves  about  AE,  the  sides  of  the 
polygon  generate  surfaces  whose  areas  are  as  follows : 

area  AB  =  AFx  2'jra.  §  762 

area  BC  =  FO  X  27ra. 
area  CD  =  OG  X  27ra. 
area  DE  =  GE X^ira. 
Adding,     area  ABCDE=  AEx  2ira. 

Now  suppose  the  number  of  sides  of  the  semi-polygon  to  be 

indefinitely  increased  ;  then  the  limit  of  the  area  ABODE  is 

the  area  of  the  surface  of  the  sphere,  and  the  limit  of  a  is  B. 

Hence  the  area  of  the  surface  of  the  sphere  =  AEx  ^ttR.  ?  260 

aE.D. 


THE   SPHERE.  369 

764.  Cor.  1.  If  /S'  denotes  the  area  of  the  surface  of  a 
sphere,  then  by  §  763, 

But  ttH^  is  the  area  of  a  great  circle ;  therefore, 

The  surface  of  a  sphere  is  equivaleni  to  four  greed  circles. 

^765.  Cor.  2.  Let  R  and  R^  denote  the  radii,  D  and  ly  the 
diameters,  and  xS'and  8'  the  areas  of  the  surfaces  of  two  spheres; 
then,  by  §  764, 

8=^'jrR',     8'  =  4:7rR'\ 

.  S  _4:7rR'  ^R'  _^(il)y  ^D' 
"8'     47ri2"     R"     (iJDy     D"' 

Therefore,  the  areas  of  the  surfaces  of  two  spheres  are  as  the 
squares  of  their  radii,  or  as  the  squares  of  their  diameters. 

766.  Cor.  3.  If  we  apply  the  reasoning  of  §  763  to  the  zone 
generated  by  the  revolution  of  the  arc  BCD,  we  obtain  for 
the  result, 

area  of  zone  BCD  =  FG  X  27rR. 

Now  FO  is  the  altitude  of  the  zone ;  therefore, 
■    The  area  of  a  zone  is  equal  to  the  product  of  its  altitude  hy 
the  drcumf&rence  of  a  great  circle. 

767.  Cor.  4.  Zones  on  the  same  sphere,  or  equal  spheres,  are 
to  each  oiher  as  their  altitudes. 

768.  Cor.  5.  The  arc  AB  generates  a  zone  of  one  base; 
and  zone  AB  =  AFx2TTR=^irAFxAE.  Now  since 
AFx  AF=  AB"  (§  337),  the  zone  AB  =  7r  AB\ 

That  is,  a  zone  of  one  hose  is  equivalent  to  a  circle  whose 
radius  is  the  chord  of  the  generating  arc. 


Ex.  583.  Find  the  area  of  the  surface  of  a  sphere  whose  radius  is  6 
inches.  -■*' .'  1    y 

Ex.  584.  Find  the  area  of  a  zone  if  its  altitude  is  3  inches,  and  the 
radius  of  the  sphere  is  6  inches.    . 


370 


V/ 


SOLID    GEOMETRY.  —  BOOK    VIII. 


Proposition  XXVIII.    Theorem. 


769,  The  area  of  a  lune  is  to  the  area  of  the  sur- 
face  of  the  sphere  as  the  number  of  degrees  in  its 
angle  is  to  S60, 


Let  ABEC  be  a  Inne,  BCDF  the  great  circle  whose 
pole  is  A;  also  let  A  denote  the  number  of  degrees  in 
the  angle  of  the  lune,  L  the  area  of  the  lune,  and  S 
the  area  of  the  surface  of  the  sphere. 

To  prove  that  Z:  S=  A:  360. 

Proof.   The  arc  ^(7  measures  the  Z  A  of  the  lune.         §  722 

Hence,  arc  £0:  circumference  £CDF=  A  :  360. 

(1)  If  BO  and  BCDF  are  commensurable,  let  their  com- 
mon measure  be  contained  tyi  times  in  ^(7,  and  n  times  in 
BCDF.     Then 

arc  BQ\  circumference  BCDF—  m  :  n. 

.'.  A:S60  =m:n.  §  262 

Pass  arcs  of  great  (D  through  the  diameter  AF  and  all  the 

points  of  the  division  of  BCDF.     These  arc^  "will  divide  the 

entire  surface  into  n  equal  lunes,  of  which  th^.  lune  ABEC 

will  contain  tti. 

.*.  L  :  8='m :  n. 

,\L:8=A:Zm. 

(2)  If  BC  and  BCDF  are  incommensurable^  the  theorem 
can  be  proved  by  the  method  of  limits  as  in  §  261.  a  e.  a 


THE   SPHERE.  371 

770.  Cor.  1.  If  L  and  S  are  expressed  as  spherical  degrees 
(§  761),  then  since  S  contains  720  spherical  degrees, 

Z:720  =  J[:360^ 
Whence  L  =  2A. 

That  is, 

The  numerical  value  of  a  lune  expressed  in  spherical  degrees  is 
twice  the  numerical  value  of  its  angle  expressed  in  angle-degrees. 

771.  Cor.  2.  If  L  and  8  are  expressed  in  ordinary  units  of 
area  (as  square  inches,  etc.),  then,  since  8~  ^irB^^ 

L-.^-nR^^A  :360°. 

Whence  ^  =  ''^- 

90° 

772.  Cor.  3.  If  we  compare  two  lunes  on  the  same  sphere, 
or  equal  spheres,  R  is  constant ;  hence,  if  Z,  V  denote  the 
lunes,  A,  A'  their  angles, 

L-n-'^EA-  E^^'-A  -A' 
^''^~    90°    •     90^~^-^- 
That  is, 

Txoo  lunes  on  the  same  sphere,  or  equal  sphet^es,  have  the 

same  ratio  as  their  angles. 

TIZ.  Cor.  4.  If  we  compare  two  lunes  X,  L\  which  have 
the  same  Z  A,  but  are  situated  on  unequal  spheres  whose 
radii  are  R  and  R\  then 

L  :  L'  --^llEA  :  I^IA  =  E' :  R'\ 
90°         90° 

Two  lunes  on  unequal  spheres  which  have  the  same  angle 

may  be  called  similar  lunes.     Therefore, 

*    Si'inilar  lunes  have  the  same  ratio  as  the  squares  of  the  radii 

of  the  spheres  on  which  they  are  situated.        v 

Ex.  585.  Given  the  radius  of  a  sphere  10  inches  ;  find  the  area  of  a 
lune  whose  angle  is  30°.  .  .-  ...       \  _, 

Ex.  586.  Given  the  diameter  of  a  sphere  16  inches ;  find  the  area  of 
a  lune  whose  angle  is  75°. 


kL    SOLID  GEOMETEY.  —  BOOK  VIII. 

\ 

Proposition  XXIX.     Theorem. 

774.    The   area  of  a  spherical  triangle,   expressed 
Mn  spherical   degrees,   is  numerically  equal   to    the 
spherical  excess  of  th&  triangle. 

.0 


c 

Let  A,  B,  C  denote  the  values  of  the  angles  of  the 
spherical  triangle  ABC,  and  E  the  spherical  excess. 

To  prove  that  the  number  of  spherical  degrees  in  A  ABC—  E. 

Proof.   Produce  the  sides  oi  /\  ABCio  complete  circles. 

These  circles  divide  the  surface  of  the  sphere  into  eight 
spherical  triangles,  of  which  any  four  having  a  common  ver- 
tex, as  A,  form  the  surface  of  a  hemisphere,  and  therefore  con- 
tain 360  spherical  degrees. 

Now  A  /li?e+ A  y1'/?(7=olune  ABA'C. 
And  the  A  A^BC,  AB'C  are  symmetrical. 

.-.AA'BC-AAB'C.  ^  §748 

.-.  A  ABC+  A  AB'C  =0=  lune  ABA'C. 
Also  A  ABC-\-  A  AB'O  =0=  lune  BAB'C.    - 

,And  A^^(7+A^i?e'  =o=lune  e^C''^.  '         ,^ 

Add  and  observe  that  in  spherical  degrees  ^ 

A  ABO-}-  AB'C  +  AB'C+  ABC'=Sm, 
and  lunes  ABA' C-{- BAB'C-}-  CAC'B  are  numerically  equal 
to  2  (^  +  ^  +  C),  and  we  have  §  770 

2  A  A B0-\-  Se>0  ^  2(A-}-  B -I-  C). 
Whence      A  A  BC^-^  A  -\-  B  +  C-  180  -  E.  q.  e.  o 


THE  SPHERE.  373 

776.   Cor.  1.   Since  in  Bpherical  degrees  A  ABC=  E,  and 
the  entire  surface  of  the  sphere  =  720,  therefore, 
A  ABC :  entire  surface  =-  E  :  720. 

That  is, 

The  area  of  a  spherical  triangle  is  to  the  area  of  the  surface 
of  the  sphere  as  the  number  which  expresses  its  spherical  excess 
is  to  720. 

776.   Cor.  2.   Hence  we  may  easily  express  the  value  of 
A  ABC  in  ordinary  units  of  area  (as  square  inches,  etc.). 
For,  let  S  denote  the  area  of  the  surface  of  the  sphere. 
Then  A  ABC:  8^  E  :  720. 

.'.AABC=4^^ 
720 

Bnt  8=  i'jrB' (^  764). 

AttB'E     it  re 


'.AABC=^ 


720  180 


Ex.  587.  What  part  of  the  surface  of  a  sphere  is  a  triangle  whose 
angles  are  120°,  100°,  and  95°  ?  "What  is  its  area  in  square  inches,  if  the 
radius  of  the  sphere  is  6  inches  ? 

Ex.  588.  Find  the  area  of  a  spherical  triangle  whose  angles  are  100°, 
120°,  140°,  if  the  diameter  of  the  sphere  is  16  inches. 

Ex.  589.  If  the  radii  of  two  spheres  are  6  inches  and  4  inches  respec- 
tively, and  the  distance  between  their  centres  is  5  inches,  what  is  the 
area  of  the  circle  of  intersection  of  these  spheres  ? 

Ex.  590.  Find  the  radius  of  the  circle  determined  in  a  sphere  of  6 
inches  diameter  by  a  plane  1  inch  from  the  centre. 

Ex,  591.  If  the  radii  of  two  concentric  spheres  are  R  and  R\  and  if 
a  plane  is  drawn  tangent  to  the  interior  sphere,  what  is  the  area  of  the 
section  made  in  the  other  sphere  ? 

Ex.  592.  W^koints  A  and  B  are  8  inches  apart.  Find  the  locus  in 
space  of  a  point^l'nnches  from  A  and  7  inches  from  B. 

Ex.  593.  The  radii  of  two  parallel  sections  of  the  same  sphere  are  a 
and  h  respectively,  and  the  distance  between  these  sections  is  d;  find 
the  radius  of  the  sphere. 


374 


SOLID    GEOMETRY. — BOOK    VIII. 


Proposition  XXX.     Theorem. 


777.  If  T  denotes  the  sum  of  the  angles  of  a 
spherieal  polygon  of  n  sides,  the  area  of  the  poly- 
gon expressed  in  spherical  degrees  is  numerically 
equal  to  T-(n-2)18(P, 


Let  ABODE  be  a  polygon  of  n  sides. 

To  prove  that  the  area  of  ABODE  is  numerically  equal  to 
2'_(^_2)180°. 

Proof,  Divide  the  polygon  into  spherical  triangles  by  draw- 
ing diagonals  from  any  vertex,  as  A. 

These  diagonals  will  divide  the  polygon  into  n  —  2  spherical 
triangles,  and  the  area  of  each  triangle  in  spherical  degrees  is 
numerically  equal  to  the  sum  of  its  angles  minus  180°.     §  774 

Hence  the  sum  of  the  areas  of  all  the  w  —  2  triangles  is  nu- 
merically equal  to  the  sum  of  all  their  angles  minus  (n  —  2)  180°. 

Now  the  sum  of  the  areas  of  the  triangles  is  the  area  of  the 
polygon,  and  the  sum  of  their  angles  is  the  sum  of  the  angles 
of  the  polygon,  that  is,  T, 

Therefore  the  area  of  the  polygon  is  numerically  equal  to 
r- (71-2)  180°.  

Ex.  594.  Find  the  area  of  a  spherical  quadrangle  whose  angles  are 
170°,  139°,  126°,  and  141°,  if  the  radius  of  the  sphere  is  10  inches. 

Ex.  595.  Find  the  area  of  a  spherical  pentagon  whose  angles  are  122° 
128°,  131°,  160°,  161°,  if  the  surface  of  the  sphere  is  150  square  feet. 

Ex.  596.  Find  the  area  of  a  spherical  hexagon  whose  angles  are  9G°, 
110°,  128°,  136°,  140°,  150°,  if  the  circumference  of  a  great  cir'^le  of  tho 
sphere  is  10  inches. 


THE   SPHERE. 


375 


The  Volume  of  a  Sphere. 

778.  A  spherical  pyramid  is  the  portion  of  a  sphere  bounded 
by  a  spherical  polygon  and  the  planes  of 
its  sides.  -^^f^^^^^^^Q 

The  centre  of  the  sphere  is  the  vertex 
of  the  pyramid. 

The  spherical  polygon  is  its  base. 
Thus,  0-ABCD  is  a  spherical  pyramid. 

779.  A  spherical  sector  is  the  portion  of 
a  sphere  generated  by  the  revolution  of  a  circular  sector  about 
any  diameter  of  the  circle  of  which  the  sector  is  a  part. 

The  base  of  a  spherical  sector  is  the  zone  generated  by  the 
arc  of  the  circular  sector.  Thus,  the 
circular  sector  AOB  revolving  about 
the  line  JfiV  generates  a  spherical  sec- 
tor whose  base  is  the  zone  generated 
by  the  arc  AB\  the  other  bounding 
surfaces  are  the  conical  surfaces  gen- 
erated by  the  radii  OA  and  OB.  The  sector  generated  by 
AOMi?,  bounded  by  a  conical  surface  and  a  zone  of  one  base 
If  00  is  perpendicular  to  OM,  the  sector  generated  by  AOC 
is  bounded  by  a  conical  surface,  a  plane  surface,  and  a  zone. 

780.  A  spherical  segment  is  a  portion  of  a  sphere  contained 
between  two  parallel  planes. 

781.  The  bases  of  a  spherical  segment  are  the  sections  made 
by  the  parallel  planes,  and  the  altitude  of  a  spherical  segment 
is  the  distance  between  its  bases, 

782.  If  one  of  the  parallel  planes  is  tangent  to  the  sphere, 
the  segment  is  called  a  segment  of  one  bo^e. 

783.  A  spherical  wedge  is  a  portion  of  a  sphere  bounded  by 
a  lune  and  two  great  semicircles. 


376 


SOLID    GEOMETEY. 


BOOK    VIII. 


Proposition  XXXI.     Theorem. 

784.  The  volume  of  a  sphere  is  equal  to  the  product 
of  the  area  of  its  surface  hy  one-third  of  its  radius. 


Let  R  be  the  radius  of  a  sphere  whose  centre  is  0, 
>S  its  surface,  and  V  its  volume. 

To  prove  V=SxiB. 

Proof.  Conceive  a  cube  to  be  circumscribed  about  the  sphere. 
Its  volume  will  be  greater  than  that  of  the  sphere,  because  it 
contains  the  sphere. 

From  O,  the  centre  of  the  sphere,  conceive  lines  to  be  drawn 
to  the  vertices  of  the  cube. 

These  lines  are  the  edges  of  six  quadrangular  pyramids, 
whose  bases  are  the  faces  of  the  cub^e,  and  whose  common  alti- 
tude is  the  radius  of  the  sphere. 

The  volume  of  each  pyramid  is  equal  to  the  product  of  its 
base  by  J  its  altitude.  Hence  the  volume  of  the  six  pyramids, 
that  is,  the  volume  of  the  circumscribed  cube,  is  equal  to  the 
area  of  the  surface  of  the  cube  multiplied  by  J  i?. 

Now  conceive  planes  drawn  tangent  to  the  sphere,  at  the 
points  where  the  edges  of  the  pyramids  cut  its  surface.  We 
shall  then  have  a  circumscribed  solid  whose  volume  will  be 
nearer  that  of  the  sphere  than  is  the  volume  of  the  circum- 
scribed cube,  because  each  tangent  plane  cuts  away  a  portion 
of  the  cube. 


THE   SPHERE.  377 

From  0  cjiiceive  lines  to  be  drawn  to  each  of  the  polyhe- 
dral angles  of  the  solid  thus  formed,  a,  h,  c,  etc. 

These  lines  will  form  the  edges  of  a  series  of  pyramids, 
whose  bases  are  the  surface  of  the  solid,  and  whose  common 
altitude  is  the  radius  of  the  sphere  ;  and  the  volume  of  each 
pyramid  thus  formed  is  equal  to  the  product  of  its  base  by  \ 
its  altitude. 

Hence  the  sum  of  the  volumes  of  these  pyramids,  that  is, 
the  volume  of  this  new  solid,  is  again  equal  to  the  area  of  its 
surface  multiplied  by  -^  i?. 

Now  this  process  of  drawing  tangent  planes  may  be  consid- 
ered as  continued  indefinitely,  and,  however  far  this  process 
is  carried,  the  volume  of  the  solid  will  always  be  equal  to  the 
area  of  its  surface  multiplied  by  \  R. 

But  the  volume  of  the  circumscribed  solid  will  approach 
nearer  and  nearer  to  that  of  the  sphere ;  and  as  the  volumes 
approach  coincidence,  the  surfaces  also  approach  coincidence. 

Hence,  Y  and  8  are  the  limits  of  the  volume  and  the  sur- 
face respectively,  of  the  circumscribed  solid. 

.'.   V=By.\R.  §260 

Q.  E.  D. 

785.  Cor.  1.  Since  B-^^-kR  (§  764),  and  R-=\I),  we 
obtain  by  substitution  the  formulas 

V=\itR\  and   V=\itD\ 

786.  Cor.  2.  The  volumes  of  iivo  spheres  are  to  each  other 
as  the  cubes  of  their  radii. 

For,  if  i?,  R'  denote  the  radii,   T^and  V  the  volumes, 
V=  i-n-R',  and   V  =  i7rR'\ 
.',  V :  V  =  iirR'  :  ^ttR"  =  R'  :  R'\ 

787.  Cor.  3.  The  volume  of  a  sphetncal pyramid  is  equal  to 
the  product  of  its  base  by  one-third  of  the  radius  of  the  sphere. 

For,  it  is  obvious  that  the  reasoning  employed  iu  §  784 
applies  equally  well  to  a  spherical  pyramid. 


378  SOLID    GEOMETRY.  —  BOOK    VIII. 

788.  Cor.  4.  The  voluine  of  a  sphericojl  sector  is  equal  to  the 
product  of  the  zone  which  forms  its  base  by  one-third  of  the 
radius  of  the  sphere. 

789.  Cor.  5.  If  R  denotes  the  radius  of  a  sphere,  C  the  cir- 
cumference of  a  great  circle,  -S'the  altitude  of  the  zone,  Zthe 
surface  of  the  zone,  and  Y  the  volume  of  the  corresponding 
sector  ;  then,  since  C=  27ri?,  and  Z=  2'irIiII,  we  have 

Proposition  XXXII.     Problem. 

790.  To  find  the  volume  of  a  spherical  segment. 

D 


B  M 


Let  AC  and  BD  be  two  semi-chords  perpendicular  to 
the  diameter  MN  of  the  semicircle  NGDM.    Let 

OM=R,  AM=a,  BM^b,  AD=a-b  =  h,  AC=-r,  BD^r'. 

Case  I.  To  find  the  volume  of  the  segment  of  one  base  gen- 
erated by  the  circular  semi-segment  ACM,  as  the  semicircle 
revolves  about  NM  as  an  axis. 

The  sector  generated  by  OCM=  ^^rRa.  §  789 

The  cone  generated  by  OCA  =  \Trr''{R  -  a).  §  672 

Hence  segment  ACM=  ^irE^a  —  ^Trr^ (B  —  a) 

=  '^(2I^a-Br'  +  ar'). 

Now  7^  =  a(2Ii  —  a)  (§  337)  ;  therefore  by  substitution, 

the  segment  ACM=^  ira'  (R-f)-  (1) 


THE   SPHERE.  379 

If  from  the  relation  r'  =  a  (2i?  —  a)  we  find  the  value  of  R, 
and  substitute  it  in  (1),  we  obtain  the  volume  in  terms  of  the 
altitude  and  the  radius  of  the  base. 

The  segment  ACM=  iirr'a  +  ^ 7^a^  (2) 

Case  II.  To  find  the  volume  of  the  segment  of  two  bases  gen- 
erated hy  the  circular  semi-segTuent  AjBDC,  as  the  semicircle 
revolves  about  NM  as  an  axis. 

Since  the  volume  is  obviously  the  difference  of  the  vol- 
umes of  the  segments  of  one  base  generated  by  the  circular 
semi-segments  ^C'Jf  and  BDM,  therefore  by  formula  (1), 

segment  ABDC=^  ira^  (r  -  -")  -  irh''  (r  -  -^ 

=  irRh{a-^b)-'^{a^^ab-\-b')  , 

o 

=  irh  [{Ra  -\-Rb)~:^  (a'  -{-ab-j-  b')]. 
Since  a-b  =  h,  a'' -2ab +  b^  =  h^ ; 

therefore  a'' +  ab -{- b^  =  h""  +  S  ab -, 

also  since    (2R-a)a  =  r\  and  (2R~b)b  =  r", 

Ra-i-Rb=^±tj:l  +  ^±^. 
2  2 

Hence 

th^Begment  ABBC^  ^h  {"^^^  +  "'  +  '>'  ~^-ab] 
L     2  2  3         J 

Jr'  +  r"  ,  A'  ,     ,      h'        ,'\ 


380  SOLID   GEOMETRY.  —  BOOH:    VIII. 


Numerical  Exercises. 

^  597.  Find  the  surface  of  a  sphere  if  the  diameter  is  (i.)  10  inches ;  (ii.) 
1  foot  9  inchefc  (iii.)  2  feet  4  inches ;  (iv.)  7  feet ;  (v.)  4.2  feet ;  (vi.)  10.5 
feet. 

^    698.   Find  the  diameter  oi  u.  sphere  if  the  surface  is  (i.)  616  squari. 
inches  ;  (ii.)  38^  square  feet ;  (iii.)  9856  square  feet. 

V  599.   The  circumference  of  a  dome  in  the  shape  of  a  hemisphere  is  6Q 
feet ;  how  maijy  square  feet  of  lead  are  required  to  cover  it  ? 

^  600.   If  the  ball  on  the  top  of  St.  Paul's  Cathedral  in  London  is  6  feet 
in  diameter,  what  would  it  cost  to  gild  it  at  7  cents  per  square  inch  ? 

"^  601.  What  is  the  numerical  value  of  the  radius  of  a  sphere  if  its  sur- 
face has  the  same  numerical  value  as  the  circumference  of  a  great  circle  ? 

^602.  Find  the  surface  of  a  lune  if  its  angle  is  30°,  and  the  total  sur- 
face of  the  sphere  is  4  square  feet. 

"^  603.  What  fractional  part  of  the  whole  surface  of  a  sphere  is  a  spher 
ical  triangle  whose  angles  are  43°  27^  81°  57^  and  114°  36^? 

604.  The  angles  of  a  spherical  triangle  are  60°,  70°,  and  80°.  The  radius 
of  the  sphere  is  14  feet.     Find  the  area  of  the  triangle  in  square  feet 

605.  The  sides  of  a  spherical  triangle  are  38°,  74°.  and  128°.  The 
radius  of  the  sphere  is  14  feet.  Find  the  area  of  the  polar  triangle  in 
square  feet. 

606.  Find  the  area  of  a  spherical  polygon  on  a  sphere  whose  radius 
is  10^  feet,  if  its  angles  are  100°,  120°,  140°,  and  160°. 

607.  The  planes  of  the  faces  of  a  quadrangular  spherical  pyramid 
make  with  each  other  angles  of  80°,  100°,  120°,  and  150° ;  and  the  length 
of  a  lateral  edge  of  the  pyramid  is  42  feet.  Find  the  area  of  its  base  in 
square  feet. 

608.  The  planes  of  the  faces  of  a  triangular  spherical  pyramid  make 
with  each  other  angles  of  40°,  60°,  and  100°,  and  the  area  of  the  base  of 
the  pyramid  is  4ir  square  feet.     Find  the  radius  of  the  sphere. 

609.  The  diameter  of  a  sphere  i;;  21  feet.  Find  the  curved  surface  of  a 
segment  whose  height  is  5  feet. 

-610.  What  is  the  area  of  a  zone  of  one  base  whose  height  is  A,  and 
the  radius  of  the  base  r  \  What  would  be  the  area  if  the  height  were 
twice  as  great  ? 


NUMEHicAL  p:xeiicises.  381 

611.  In  a  sphere  whose  radius  is  r,  find  the  height  of  a  zone  whose 
area  is  equal  to  that  of  a  great  circle. 

""  612.  The  altitude  of  the  torrid  zone  is  about  3200  miles.  Find  its 
area  in  square  miles,  assuming  the  earth  to  be  a  sphere  with  a  radius  of 
4000  miles. 

"^613.  A  plane  divides  the  surface  of  a  sphere  of  radius  r  into  two 
zones,  such  that  the  surface  of  the  greater  is  a  mean  proportional  between 
the  entire  surface  and  the  surface  of  the  smaller.  Find  the  distance  of 
the  plane  from  the  centre  of  the  sphere. 

.  614.  If  a  sphere  of  radius  r  is  cut  by  two  planes  equally  distant  from 

the  centre,  po  that  the  area  of  the  zone  comprised  between  the  planes  is 
equal  to  the  sum  of  the  areas  of  its  bases,  find  the  distance  of  either  plane 
from  the  centre. 

615.  Find  the  area  of  the  zone  generated  by  an  arc  of  30°,  of  which 
the  radius  is  r,  and  which  turns  around  a  diameter  passing  through  one 
of  its  extremities. 

"^  616.  Find  the  area  of  the  zone  of  a  sphere  of  radius  r,  illuminated  by 
a  lamp  placed  at  the  distance  h  from  the  sphere. 

*^  617.  How  much  of  the  earth's  surface  would  a  man  see  if  he  were 
raised  to  the  height  of  the  radius  above  it? 

^^618.  To  what  height  must  a  man  be  raised  above  the  earth  in  order 
that  he  may  see  one-sixth  of  its  surface  ? 

^19.  Two  cities  are  200  miles  apart.  To  what  height  must  a  man 
ascend  from  one  city  in  order  that  he  may  see  the  other,  supposing  the 
circumference  of  the  earth  to  be  25,000  miles  ? 

^^  620.  Find  the  volume  of  a  sphere  if  the  diameter  is  (i.)  13  inches ;  (ii.) 
3  feet  6  inches ;  (iii.)  10  feet  6  inches  ;  (iv.)  17  feet  6  inches ;  (v.)  14,7 
feet ;  (vi.)  42  feet. 

V 

621.  Find  the  diameter  of  a  sphere  if  the  volume  is  (i.)  75  cubic  feet 
1377  cubic  inches;  (ii.)  179  cubic  feet  1152  cubic  inches  ;  (iii.)  1047.816 
cubic  feet ;  (iv.)  38.808  cubic  yards. 

""  622.    Find  the  volume  of  a  sphere  whose  circumference  is  45  feet. 

"^  623.   Find  the  volume  F  of  a  sphere  in  terms  of  the  circumference  0 
of  a  great  circle. 
'^  624.    Find  the  radius  r  of  a  sphere,  having  given  the  volume  F. 

^625.  Find  the  radius  r  of  a  sphere,  if  its  circumference  and  its  volume 
have  the  same  numerical  value. 


382  SOLID   GEOMETRY.  —  BOOK   VIII. 

^^26.  If  an  iron  ball  4  inches  in  diameter  weighs  9  pounds,  what  is 
the  weight  of  a  hollow  iron  shell  2  inches  thick,  whose  external  diameter 
is  20  inches  ? 

^  627.  The  radius  of  a  sphere  is  7  feet ;  what  is  the  volume  of  a  wedge 
whose  angle  is  36°  ? 

^  628.  What  is  the  angle  of  a  spherical  wedge,  if  its  volume  is  one  cubic 
foot,  and  the  volume  of  the  entire  sphere  is  6  cubic  feet  ? 

^  629.  What  is  the  volume  of  a  spherical  sector,  if  the  area  of  the  zone 
which  forms  its  base  is  3  square  feet,  and  the  radius  of  the  sphere  is  1 
foot? 

^  630.  The  radius  of  the  base  of  the  segment  of  a  sphere  is  16  inches, 
and  the  radius  of  the  sphere  is  20  inches  ;  find  its  volume. 

^631.  The  inside  of  a  wash-basin  is  in  the  shape  of  the  segment  of  a 
sphere ;  the  distance  across  the  top  is  16  inches,  and  its  greatest  depth  is 
6  inches  ;  find  how  many  pints  of  water  it  will  hold,  reckoning  7j  gal- 
lons to  the  cubic  foot. 

632.  What  is  the  height  of  a  zone,  if  its  area  is  S,  and  the  volume  of 
the  sphere  to  which  it  belongs  is  F? 

^  633.  The  radii  of  the  bases  of  a  spherical  segment  are  6  feet  and  8 
feet,  and  its  height  is  3  feet ;  find  its  volume. 

634.  Find  the  volume  of  a  triangular  spherical  pyramid  if  the  angles 
of  the  spherical  triangle  which  forms  its  base  are  each  100°,  and  the 
radius  of  the  sphere  is  7  feet. 

635.  The  circumference  of  a  sphere  is  28 tt  feet:  find  the  volume  of 
that  part  of  the  sphere  included  by  the  faces  of  a  trihedral  angle  at  the 
centre,  the  dihedral  angles  of  which  are  80°,  105°,  and  140°. 

636.  The  planes  of  the  faces  of  a  quadrangular  spherical  pyramid 
make  with  each  other  angles  of  80°,  100°,  120°,  and  150°,  and  a  lateral 
edge  of  the  pyramid  is  3 J  feet ;  find  the  volume  of  the  pyramid. 

'^  637.  The  radius  of  the  base  of  the  segment  of  a  sphere  is  40  feet,  and 
its  height  is  20  feet ;  find  its  volume, 

638.  Having  given  the  volume  V,  and  the  height  h,  of  a  spherical 
segment  of  one  base,  find  th%  radius  r  of  the  sphere. 

^  639.  Find  the  weight  of  a  sphere  of  radius  r,  which  floats  in  a  liquid 
of  specific  gravity  s,  with  one-fourth  of  its  surface  above  the  surface  of 
the  liquid.  The  weight  of  a  floating  body  is  equal  to  the  weight  of  the 
liquid  displaced.  , 


MISCELLANEOUS   EXERCISES.  383 


Miscellaneous  Exercises. 

'^  640.  Determine  a  point  in  a  given  plane  such  that  the  difference  of 
its  distances  from  two  given  points  on  opposite  sides  of  the  plane  shall 
be  a  maximum. 

641.  In  any  warped  quadrilateral,  that  is,  one  whose  sides  do  not  all 
lie  in  the  same  plane,  the  middle  points  of  the  sides  are  the  vertices  of  a 
parallelogram. 

"  642.  In  any  trihedral  angle,  the  three  planes  bisecting  the  three  dihe- 
dral angles  intersect  in  the  same  straight  line. 

^  643.  To  draw  a  line  through  the  vertex  of  any  trihedral  angle,  mak- 
ing equal  angles  with  its  edges. 

^  644.  In  any  trihedral  angle,  the  three  planes  passed  through  the 
edges  and  the  respective  bisectors  of  the  opposite  face  angles  intersect 
in  the  same  straight  line. 

645.  In  any  trihedral  angle,  the  three  planes  passed  through  the 
bisectors  of  the  face  angles,  and  perpendicular  to  these  faces  respec- 
tively, intersect  in  the  same  straight  line. 

"  646.  In  any  trihedral  angle,  the  three  planes  passed  through  the 
edges,  perpendicular  to  the  opposite  faces  respectively,  intersect  in  the 
same  straight  line. 

647.  In  a  tetrahedron,  the  planes  passed  through  the  three  lateral 
edges  and  the  middle  points  of  the  sides  of  the  base  intersect  in  a 
straight  line. 

"^48.  The  lines  joining  each  vertex  of  a  tetrahedron  with  the  point  of 
intersection  of  the  medial  lines  of  the  opposite  face  all  meet  in  a  point 
called  the  centre  of  gravity,  which  divides  each  line  so  that  the  shorter 
segment  is  to  the  whole  line  in  the  ratio  1 :  4. 

649.  The  straight  lines  joining  the  middle  points  of  the  opposite 
edges  of  a  tetrahedron  all  pass  through  the  centre  of  gravity  of  the 
tetrahedron,  and  are  bisected  by  the  centre  of  gravity. 

""  650.  The  plane  which  bisects  a  dihedral  angle  of  a  tetrahedron  divides 
the  opposite  edges  into  segments  proportional  to  the  areas  of  the  faces 
including  the  dihedral  angle. 

^51.  The  altitude  of  a  regular  tetrahedron  is  equal  to  the  sum  of  the 
four  perpendiculars  let  fall  from  any  point  within  it  upon  the  four 
faces. 


WD 


884  SOLID   GEOMETRY.  —  BOOB!   VIII. 


652.  Within  a  given  tetrahedron,  to  find  a  point  such  that  the  planes 
passed  through  this  point  and  the  edges  of  the  tetrahedron  shall  divide 
the  tetrahedron  into  four  equivalent  tetrahedrons. 

653.  To  cut  a  cube  by  a  plane  so  that  the  section  shall  be  a  regular 
.aexagon. 

"  654.   To   cut  a  tetrahedral   angle   so   that  the  section   shall   be   a 
parallelogram. 

655.  The  portion  of  a  tetrahedron  cut  off  by  a  plane  parallel  to  any 
fac«  is  a  tetrahedron  similar  to  the  given  tetrahedron. 

656.  Two  tetrahedrons,  having  a  dihedral  angle  of  one  equal  to  a 
dihedral  angle  of  the  other,  and  the  faces  including  these  angles  respec- 
tively similar,  and  similarly  placed,  are  similar. 

657.  Two  polyhedrons  composed  of  the  same  number  of  tetrahedrons, 
similar  each  to  each  and  similarly  placed,  are  similar. 

658.  If  the  homologous  faces  of  two  similar  pyramids  are  respec- 
tively parallel,  the  straight  lines  which  join  the  homologous  vertices  of 
the  pyramids  meet  in  a  point. 

-^659.   Two  symmetrical  tetrahedrons  are  equivalent. 

660.  Two  symmetrical  polyhedrons  may  be  decomposed  into  the  same 
number  of  tetrahedrons  symmetrical  each  to  each. 

661.  Two  symmetrical  polyhedrons  are  equivalent. 

662.  If  a  solid  has  two  planes  of  symmetry  perpendicular  to  each 
other,  the  intersection  of  these  planes  is  an  axis  of  symmetry  of  the  solid. 

663.  If  a  solid  has  three  planes  of  symmetry  perpendicular  to  each 
other,  the  three  intersections  of  these  planes  are  three  axes  of  symmetry 
of  the  solid ;  and  the  common  intersection  of  these  axes  is  the  centre  of 
symmetry  of  the  solid. 

664.  The  volume  of  a  right  circular  cylinder  is  equal  to  the  product 
of  the  lateral  area  by  half  the  radius. 

665.  The  volume  of  a  right  circular  cylinder  is  equal  to  the  product 
of  the  area  of  the  rectangle  which  generates  it,  by  the  length  of  the  cir- 
cumference generated  by  the  point  of  intersection  of  the  diagonals  of  the 
rectangle. 

666.  If  the  altitude  of  a  right  circular  cylinder  is  equal  to  the  diam- 
eter of  the  base,  the  volume  is  eqnal  to  the  total  area  multiplied  by  a 
third  of  the  radius. 


MISCELLANEOUS   EXERCISES.  386 

Construct  a  spherical  surface  with  given  radius : 

667.  Passing  through  three  given  points. 

668.  Passing  through  two  given  points  and  tangent  to  a  given  plane. 

669.  Passing  through  two  given  points  and  tangent  to  a  given 
sphere. 

670.  Passing  through  a  given  point  and  ^ngent  to  two  given  planes. 

>    671.   Passing  through    a  given   point  and   tangent  to    two   given 
spheres. 

672.  Passing  through  a  given  point  and  tangent  to  a  given  plane 
and  a  given  sphere. 

673.  Tangent  to  three  given  planes. 

674.  Tangent  to  three  given  spheres. 

675.  Tangent  to  two  given  planes  and  a  given  sphere. 

676.  Tangent  to  two  given  spheres  and  a  given  plane. 

'  677.   Find  the  area  of  a  solid  generated  by  an  equilateral  triangle 
turning  about  one  of  its  sides,  if  the  length  of  the  side  is  a. 

678.  Find  the  centre  of  a  sphere  whose  surface  shall  pass  through 
three  given  points,  and  shall  touch  a  given  plane. 

679.  Find  the  centre  of  a  sphere  whose  surface  shall  touch  two  given 
planes,  and  also  pass  through  two  given  points  which  lie  between  the 
planes. 

680.  Through  a  given  point  to  pa.ss  a  plane  tangent  to  a  given  cir- 
cular cylinder. 

681.  Through  a  given  point  to  pass  a  plane  tangent  to  a  given  cir- 
cular cone. 

682.  Through  a  given  straight  line  without  a  given  sphere,  to  pass  a 
plane  tangent  to  the  sphere. 

683.  The  volume  of  a  sphere  is  two-thirds  of  the  volume  of  a  circum- 
scribing cylinder,  and  its  surface  is  two-thirds  of  the  total  surface  of  the 
cylinder. 

684.  Given  a  sphere,  a  cylinder  circumscribed  about  the  sphere,  and 
a  cone  of  two  nappes  inscribed  in  the  cylinder ;  if  any  two  planes  are 
drawn  perpendicular  to  the  axis  of  the  three  figures,  the  spherical  seg- 
ment between  the  planes  is  equivalent  to  the  difference  between  the 
corresponding  cylindrical  and  conic  segments. 


386  SOLID   GEOMETRY.  —  BOOK   VIII. 

685.  Compare  the  volumes  of  the  solids  generated  by  a  rectangle 
turning  successively  about  two  adjacent  sides,  the  lengths  of  these  sides 
being  a  and  h. 

686.  An  equilateral  triangle  revolves  about  one  of  its  altitudes. 
Compare  the  convex  surface  of  the  cone  generated  by  the  triangle  and 
the  surface  of  the  sphere  generated  by  the  circle  inscribed  in  the 
triangle. 

687.  An  equilateral  triangle  revolves  about  one  of  its  altitudes. 
Compare  the  volumes  of  the  solids  generated  by  the  triangle,  the 
inscribed  circle,  and  the  circumscribed  circle. 

688.  The  perpendicular  let  fall  from  the  point  of  intersection  of  the 
medial  lines  of  a  given  triangle  upon  any  plane  not  cutting  the  triangle 
is  equal  to  one-third  the  sum  of  the  perpendiculars  from  the  vertices  of 
the  triangle  upon  the  same  plane. 

689.  The  perpendicular  from  the  centre  of  gravity  of  a  tetrahedron 
upon  any  plane  not  cutting  the  tetrahedron  is  equal  to  one-fourth  the 
sum  of  the  perpendiculars  from  the  vertices  of  the  tetrahedron  upon  the 
same  plane. 

690.  The  volume  of  any  polyhedron  having  for  its  bases  any  two 
polygons  whose  planes  are  parallel,  and  for  lateral  faces  trapezoids,  is 
the  product  of  one-sixth  the  distance  between  the  bases  into  the  sum  of 
the  two  bases  plus  four  times  a  section  midway  between  the  bases ;  that 
is,  if  H  denotes  the  distance  between  the  bases  B  and  6,  and  B^  a  section 
midway  between  the  bases, 

V=\H{B-\-h  +  ^B^\ 

Note.  From  any  point  0  in  the  section  midway  between  the  bases, 
draw  lines  to  the  vertices  of  the  solid  angles  of  the  polyhedron,  thus  divid- 
ing the  solid  into  pyramids.  The  pyramids  having  B  and  h  as  bases,  evi- 
dently equal  \H{B-\-  h).  It  remains  to  be  proved  that  the  volume  of 
each  pyramid  having  a  lateral  face  as  its  base  equals  \  H  into  four  times 
that  portion  of  the  section  midway  between  the  bases  intercepted  by  thie 
pyramid.    This  theorem  is  much  used  in  earth-work. 


BOOK   IX. 
CONIC     SECTIONS. 


The  Parabola. 

(\J9^  The  curve  traced  by  a  point  which  moves  so  that  its 
distance  from  a  fixed  point  is  always  equal  to  its  distance 
from  a  fixed  line  is  called  a  parabola.  The  curve  lies  in  the 
plane  of  the  fixed  point  and  line. 

792.  The  fixed  point  is  called  the  foctis;  and  the  fixed  line, 
the  directnx. 

793.  A  parabola  may  be  described  by  the  continuous  motion 
of  a  point,  as  lollows : 

E 


Place  a  ruler  so  that  one  of  its  edges  shall  coincide  with 
the  directrix  DE.  Then  place  a  right  triangle  with  its  base 
edge  in  contact  with  the  edge  of  the  ruler.  Fasten  one  end 
of  a  string,  whose  length  is  equal  to  the  other  edge  BO,  to  the 
point  B,  and  the  other  end  to  a  pin  fixed  at  the  focus  F, 
Then  slide  the  triangle  BCE  along  the  directrix,  keeping  the 
string  tightly  pressed  against  the  ruler  by  the  point  of  a  pen- 
cil P.  The  point  P  will  describe  a  parabola ;  for  during  the 
motion  we  always  have  PF=  PC. 


388 


GEOMETRY.  —  BOOK   IX. 


Proposition  I.     Problem. 

,£9p  To  construct    a    -parabola    by    points,   having 
given  its  focus  and  its  directri.v. 


Let  F  be  the  focus,  and  CDE  the  directrix. 

To  construct  the  parabola  by  points. 

Construction.    Draw  FD  ±  to  CE,  and  meeting  CE  at  D. 

Bisect  FD  at  A.     Then  ^  is  a  point  of  the  curve.       §  791 

Through  any  point  M  in  the  line  DF,  to  the  right  of  A, 
draw  a  line  II  to  CE. 

With  i^as  centre  and  EM SiS  radius,  draw  arcs  cutting  this 
line  at  the  points  P  and  Q. 

Then  P  and  Q  are  points  of  the  curve. 

Proof.  Draw  FC,  QE±  to  CE. 

Then  FC=  EM,  and  QE=  EM, 

and  EM=FF=QF. 

.-.  PC^PF;  and  QE  =  QF. 
Therefore  P  and  Q  are  in  the  curve. 

In  this  way  any  number  of  points  may  be  found  ;  and  a 
continuous  curve  drawn  through  the  points  thus  determined 
will  be  the  parabola  whose  focus  is  Pand  directrix  CDE. 

Q.  E.  F. 


Cons. 


§791 


THE   PARABOLA.  389 

795.  The  point  A  is  called  the  vertex  of  the  curve.  The 
line  DF  produced  indefinitely  in  both  directions  is  called  the 
axis. 

796.  The  line  FF,  joining  the  focus  to  any  point  F  on 
the  curve,  is  called  the  focal  radius  of  F. 

797.  The  distance  AM  ia  called  the  abscissa,  and  the  dis- 
tance FM  the  ordinate,  of  the  point  F. 

798.  The  double  ordinate  LF,  through  the  focus,  is  called 
the  latus  rectum  or  parameter. 

799.  Cor.  1.  Since  FF=  FQ  (Cons.),  MF=  MQ  (§  121)  ; 
hence,  the  parabola  is  symmetrical  with  respect  to  its  axis 
(§63). 

800.  Cor.  2.  The  curve  lies  entirely  on  one  side  of  the  per- 
pendicular to  the  aods  ei'ected  at  the  vertex ;  namely,  on  the 
same  side  as  the  focus. 

For,  any  point  on  the  other  side  of  this  perpendicular  is 
obviously  nearer  to  the  directrix  than  to  the  focus. 

801.  Cor.  3.    The  parabola  is  not  a  closed  curve. 

For  any  point  on  the  axis  of  the  curve  to  the  right  of  F  is 
evidently  nearer  to  the  focus  than  to  the  directrix.  Hence 
the  parabola  QAF  cannot  cross  the  axis  to  the  right  of  F. 

802.  Cor.  4.    The  latus  rectum  is  equal  to  4:AF. 
For,  draw  LG  ±  to  OFF. 

Then,  LF=  LG,  and  ZG  =  FF. 

.'.LF=FF=2AF. 
Similarly,  FF=  FF=2AF 

Therefore  LF-=4:AF 


803.  Remark.  In  the  following  propositions,  the  focus  will  be 
denoted  by  F,  the  vertex  by  A,  and  the  point  where  the  axis  meets  the 
directrix  by  D. 


390 


GEOMETEY.  —  BOOK    IX. 


Proposition  II.     Theorem. 

804.  The  ordinate  of  any  point  of  a  parabola  is  a 
mean  proportional  between  the  latus  rectum  and  the 
abscissa. 


Let  P  be  any  point,  AM  its  abscissa,  PM  its  ordinate. 

-2 


To  prove 
Proof. 


PM  =4:AFxAM. 


§791 


Hence 


Q.  E.  D. 


iW  =  FP  -FM"  =  Dlt  -  FM" 
=  {DM-  FM)  {DM+  FM) 
=  DF{DF+  FM-\-  FM) 
=  2AF{2AF-\-2FM) 
~2AF{2AM). 

FM'  =  4:AFxAM.  (1) 

805.  Cor.  1.  The  (greater  the  abscissa  of  a  point,  the  greater 
the  ordinate.     For  PJIf  increases  with  AM  in  equation  (1), 

806.  Cor.  2.  If  P  and  Q  are  any  two  points  of  the  curve, 

FM'_4:AFx  AM_AM 

Q^'    4:AFxan    an' 

Hence,  the  squares  of  any  two  ordinatcs  are  as  the  abscissas. 


THE   PARABOLA. 


391 


Proposition  III.     Theorem. 

807.  Every  point  within  the  parabola  is  nearer  to 
the  focus  than  to  the  directrix;  and  every  point  with- 
out the  parabola  is  farther  from  the  focus  than  from 
the  directrix. 


1.  Let  Q  be  a  point  within  the  parabola.  Draw  QC 
perpendicular  to  the  directrix,  cutting  the  curve  at 
P.   Draw  QF,  PF, 

To  prove  QF  <  QC. 

Proof.   In  the  A  QPF,  QF<QP-\-  PF.  §  137 

But  PF=  PC.  §  791 

:.QF<  QP-\-PC, 

.:QF<QC 

2,  Let  Q'  be  a  point  without  the  curve.    Draw  Q'F. 

To  prove  Q'F>  Q'C 

Proof.  In  the  A  Q'FP,  Q'F  >  PF-  PQ\  §  137 

or  QF>  PC~PQ. 

That  is,  Q'F>Q'a  ae-D. 

808.  Cor.  A  point  is  within  or  without  a  parabola  according 
as  its  distance  from  the  focus  is  less  than,  or  greater  than,  its 
distance  froon  the  directrix. 

809.  A  straight  line  which  touches,  but  does  not  cut,  a 
parabola,  is  called  a  tangent  to  the  parabola.  The  point 
where  it  touches  the  parabola  is  called  the  point  of  contact. 


392 


GEOMETRY. 


BOOK    IX. 


Proposition  IV.     Theorem. 

810.  If  a  line  PT  is  drawn  from  any  point  P  of 
the  curve,  bisecting  the  angle  between  FF  and  the 
perpendicular  from  F  to  the  directrix,  every  point  of 
the  line  FT,  except  F,  is  without  the  curve,  j^ 
H 


Let  PG  be  the  perpendicular  from  P  to  the  directrix, 
the  angle  FPT  equal  the  angle  OPT,  and  let  X  he  any- 
other  point  in  PT  except  P. 

To  prove  that  X  is  without  the  curve. 

Proof.  Draw  XE  ±  to  the  directrix,  and  join  CX,  FX,  OF, 
and  let  OF  meet  P^at  R. 

In  the  isos.  A  FCF,       OR  =  RF.  Ex.  14 

Hence  CX=  FX.  §  122 

But  EX<  ex.  §  114 

Therefore  EX  <  FX. 

That  is,  JTis  without  the  curve.  §  808 

Q.  E.  D. 

811.  Cor.  1.  FT  is  the  tangent  at  the  point  F  (§  809). 

812.  Cor.  2.  FThisects  EC,  and  is  perpendicular  to  EG. 

813.  Cor.  3.  Since  the  angles  EFT  and  FTP  are  equal, 
FT  equals  EF(I  156). 


THE   PARABOLA.  393 

814.   Cor.  4.   The  tangent  at  A  is  pe^'pcndicular  to  the  axis. 
For  it  bisects  the  straight  angle  FAD. 

315.   Cor.  5.   The  tangent  at  A  is  the  locus  of  the  foot  of  the 
perpendicular  dropped  from  the  focus  to  any  tangent. 
Since  FR  =  EC,  and  FA  =  AD,  B,  is  in  AR{^  811). 

816.  The  line  FN  drawn  through  F  perpendicular  to  the 
tangent  PTis  called  the  normal  at  F. 

817.  If  the  ordinate  of  F  meet  the  axis  in  M,  and  the  tan- 
gent and  normal  at  Pmeet  the  axis  in  T  and  JV  respectively, 
then  MT'\^  the  suhtangent  and  MN  ih.Q  subnormal. 

^  818.  Cor.  6.  The  suhtangeni  is  bisected  by  the  vertex. 

For,  FT=  FF,                               §  813 

and  FF^DM.                              §791 

Hence  FT=  DM',  also  AF=  AD. 

Therefore  FT-  AF=  DM-  AD, 

or  TA  =  AM. 

V    819.   Cor.  7.   The  subnormal  is  equal  to   half  the  IcUus 
rectum. 

For  CF  =  FN,  and  CF  =  DM.                    §  ISO 

Hence  PiV-  DM, 

or  FM-\-  ¥JSr=  DF-\-  FM. 

Therefore  MN=  DF. 

820.  Cor.  8.  The  normal  bisects  the  angle  between  FF  and 
CF  produced;  that  is,  bisects  the  angle  FFO. 

For  Z  NFT=  Z  NFK,  and  Z  FFT=  Z  TFC=-  Z  OFK. 
Hence  Z  NFF=  Z  NFG. 

821.  Cor.  9.  The  circle  with  F  as  centre  and  FF  as  radius 
parses  through  T  and  N.  / 


394 


GEOMETRY.  —  BOOK    IX. 


Proposition  V.     Problem. 

822.  To  draw   a  tangent  to  a  parabola  from  an 
exterior  point. 


Let  B  be  any  point  exterior  to  the  parabola,  QAP. 

To  draw  a  tang e7it  from  M  to  QAP. 

Oonstmction.  With  R  as  centre  and  RF  as  radius,  draw 
arcs  cutting  the  directrix  at  the  points  B,  C.  Through  ^  and 
Cdraw  lines  parallel  to  the  axis,  and  meeting  the  parabola  in 
P,  Q,  respectively.  Join  RP,  RQ.  Then  RP  and  RQ  are 
tangents  to  the  curve. 

Proof.  RB  =  RF,  Cons. 

PB  =  PF.  §  791 

Hence  Z  RPB  =  Z.  RPF.  §  160 

Therefore  RP  is  the  tangent  at  P.  §  811 

For  like  reason,  RQ  is  the  tangent  at  Q.  atF. 

823.  Cor.  Since  R  is  without  the  curve,  it  is  nearer  to  the 
directrix  than  to  the  focus  (§  807);  therefore,  the  circle  with  R 
as  centre  and  RF  as  radius,  must  always  cut  the  directrix  in 
two  points  ;  therefore,  two  tangents  can  always  he  drawn  to  a 
parabola  from  an  exterior  point. 

824.  The  line  joining  the  points  of  contact  P  and  Q  is  called 
the  chord  of  contact  for  the  tangents  drawn  from  R. 


THE    PARABOLA.  395 


Proposition  VI.     Theorem. 

825.  The  line  joining  the  foeus  to  the  intersection 
of  two  tangents  rnakes  equal  angles  with  the  focal 
radii  drawn  to  the  points  of  contact. 

Let  the  tangents  drawn  at  P  and  Q  meet  in  R. 

To  prove  Z  RFP  =  Z  BFQ. 

Proof,        Draw  the  Js  FjB,  QC  to  the  directrix, 
and  join  FB,  FC,  FF. 


Since 

FB  =  FF, 

and 

FF=FF, 

§§  812,  112 

AFFF^AFBF, 

§160 

and 

ZFFF^ZFBF. 

Similarly, 

Z  FFQ  -  Z  FCQ. 

Now, 

ZFBF=90''+ZFBC, 

and 

Z  FCQ  =  90° i-Z  FOB; 

and  since 

FB  =  FF,mdFC=FF, 

therefore 

FB  =  FC. 

Hence 

ZFBC^ZFCB. 

§154 

Therefore 

ZFBF=ZFCQ, 

and 

ZFFF=ZFFQ. 

aE.D. 

826.   Cor.  If  the  chord  of  contact  FQ  passes  through  F, 
then  FFQ  is  a  straight  line. 

Hence  FFF  +  FFQ  =  180°, 

and  FFF=FFQ  =  90\ 

Therefore  FBF  =  FCQ  =  90°. 

Therefore,  the  tangents  drawn  through  the  ends  of  a  focal 
chord  meet  in  the  directrix. 


396 


GEOMETRY. 


BOOK   IX. 


Proposition  VII.     Theorem. 

827.  If  a  pair  of  tangents  are  drawn  from  a  point 
R  to  a  parabola,  the  line  drawn  through  R  parallel 
io  the  axis  will  hiseet  the  chord  of  contact. 


Let  the  tangents  drawn  from  li  meet  the  curve  in 
P,  Q,  and  let  the  line  through  R  parallel  to  the  a:xis 
meet  the  directrix  in  H,  the  curve  in  S,  and  the  chord 
of  contact  in  M. 

To  prove  FM=  QM. 

Proof.        Drop  the  Js  FR,  QC  to  the  directrix, 
and  join  RR,  RC. 

RJIis±  to  RO,  §102 

RR^RC.  .    .  §823 

Hence  RJ3'=  CH.  §  121 

Since  RR,  QC,  and  i^Jf  are  II,  §  100 

therefore  RM=  QM.  §  187 

Q.  E.  D 


THE    PARABOLA.  397 

Proposition  VIII.     Theorem. 

828.  //  a  pair  of  tangents  EP,  RQ  are  drawn  from 
a  point  B  to  a  parabola,  and  through  E  a  lUie  par- 
allel to  the  axis  is  drawn,  meeting  the  curve  in  8, 
the  tangent  at  8  will  he  parallel  to  the  chord  of  con- 
tact. 

Let  the  tangent  at  S  meet  the  tangents  PR,  QR  in 
T,  V,  respectively. 

To  prove  TV  W  to  FQ. 

Proof.   Draw  TN  II  to  8M,  and  let  it  meet  8F  iii  N. 

Then  Pi\r=  ]^8  §  827 

Hence  FT=  TE.  §  188 

Similarly,  QV=  VE. 

Therefore  TV  is  II  to  FQ.  §  189 

aE.  D. 

^  829.  Cor.  1.  If  we  suppose  E  to  move  along  E3f  towards 
the  curve,  then  since  the  point  8  and  the  direction  of  the  tan- 
gent TF" remain  fixed,  the  chord  FQ  will  remain  parallel  to 
TV,  while  its  middle  point  M  will  move  along  i?Jf  towards 
8 ;  finally,  E,  M,  P,  and  Q  will  all  coincide  at  xS'. 

Hence,  the  line  EM  is  the  bctis  of  the  middle  points  of  all 
cho7'ds  drawn  parallel  to  the  tangent  at  8. 

830.  The  locus  of  the  middle  points  of  a  system  of  parallel 
chords  in  a  parabola  is  called  a  diameter.  The  parallel  chords 
are  called  the  ordinates  of  the  diameter. 

831.  Cor.  2.  The  diameters  of  a  parabola  are  parallel  to  its 
axis ;  and  conversely,  efoefry  straight  line  parallel  to  the  axis  is 
a  diameter ;  that  is,  bisects  a  system  of  parallel  chords. 

832.  CoR.  3.  Tangents  drawn  through  the  ends  of  an  ordi- 
nate intersect  in  the  diameter  corresponding  to  that  ordinate. 

833.  CoR.4.  The  point /S' is  the  middle  point  of  Pif(§  188); 
therefore,  the  portion  of  a  diameter  contained  between  any  ordi- 
nate and  the  intersection  of  the  tangents  drawn  through  the  ends 
of  the  ordinate  is  bisected  by  the  curve. 


398 


GEOMETRY. 


BOOK    IX. 


834.  Cor.  5.  The  point  S  is  also  the  middle  point  of  the 
tangent  TV;  therefore,  the  part  of  a  tangent  parallel  to  a 
chord  contained  between  the  two  tangents  drawn  through  the 
ends  of  the  chord  is  bisected  by  the  diameter  of  the  chord  at 
the  point  of  contact. 

Proposition  IX.     Theorem. 

835.  The  area  of  a  parabolic  segment  made  hy  a 
chord  is  two-thirds  the  area  of  the  triangle  formed 
hy  the  chord  and  the  tangents  drawn  through  the 
ends  of  the  chord. 

X  


Ti 


m 


Q 

Let  PQ  be  any  chord,  and  let  the  tangents  at  P  and 
Q  meet  in  R. 

To  prove  segment  P8Q  =  |  A  PRQ. 


THE    PARABOLA.  399 

Proof.   Draw  the  diameter  RM,  meeting  the  curve  at  8,  and 

at  8  draw  a  tangent  meeting  PR  in  T  and  QR  in  V.     Join 
8P,  8Q. 

Since                 FT=  TR,  and  qy=.  VR,    "^^  §  828 
FT  is  II  to  PQ, 

and                                  PQ  =  2x  VT.  §  189 

.-.  A  PQ8=  2  A  TFi?.  §  370 

If  now  we  draw  through  T,  V,  the  diameters  T8',  V8",  and 
then  draw  through  8',  8",  the  tangents  T'8'V',  T"8"V",  we 
can  prove  in  the  same  way  that 

AP88'  =  2AT'V'T, 
and  AQ88"=2AT"V"V. 

If  we  continue  to  form  new  triangles  by  drawing  diameters 
through  the  points  T',  V\  T'\  F",  and  tangents  at  the  points 
where  these  diameters  meet  the  curve,  we  can  prove  that  each 
interior  triangle  formed  by  joining  a  point  of  contact  to  the 
extremities  of  a  chord  is  twice  as  large  as  the  exterior  triangle 
formed  by  the  tangents  through  these  points.  '  And  this  is 
true  however  long  the  process  is  continued. 

Therefore  the  sum  of  all  the  interior  triangles  is  equal  to 
twice  the  sum  of  the  corresponding  exterior  triangles. 

Now  if  we  suppose  the  process  to  be  continued  indefinitely, 
then  the  limit  of  the  sum  of  the  interior  triangles  will  be  the 
area  contained  between  the  chord  PQ  and  the  curve,  and  the 
limit  of  the  sum  of  the  exterior  triangles  will  be  the  area  con- 
tained between  the  curve  and  the  tangents  PR,  QR. 

Hence  segment  PQ8=  twice  the  area  contained  by  PR, 
QR,  and  the  curve,*       =  |  A  PQR.  §  260 

aE.D. 

836.  Cor.  If  throtcgh  P  and  Q  lines  are  drawn  parallel  to 
8M,  meeting  the  tangent  TV  produced  in  the  points  X  and  Y, 
then-the  segment  PQ8=  i  O  PQ  YX. 


400 


GEOMETRY.  —  BOOK   IX. 


Proposition  X.     Theorem. 

837.  The  section  of  a  right  circular  cone  made  hy  a 
-plane  -parallel  to  one,  and  only  one,  element  of  the 
surface  is  a  parabola. 


Let  SB  be  any  element  of  the  cone  whose  axis  is 
8Z,  and  let  QAP  be  the  section  of  the  cone  made  by  a 
plane  perpendicular  to  the  plane  BSZ  and  parallel 
to  SB. 

To  prove  that  the  curve  PAQ  is  a  parabola. 

Proof.  Let  SO  he  the  second  element  in  whicli  the  plane 
BSZ  cuts  the  cone,  and  let  BAD  be  the  intersection  of  the 
planes  ^xS'^ and  PAQ. 

Draw  the  O  0  tangent  to  the  lines  /^,  SO,  BD,  and  let 
G,  H,  F  be  the  points  of  contact  respectively. 

Kevolve  BSQ  and  the  O  OOH  about  the  axis  SZ,  the 
plane  PAQ  remaining  fixed.  The  O  0  will  generate  a  sphere 
which  will  touch  the  cone  in  the  O  GKII,  and  the  plane  PA  Q 
at  the  point  F. 


THE   PARABOLA.  401 

Since  80  is  ±  to  GJE,  SO  is  ±  to  the  plane  GKR.     §  462 

Hence  the  plane  JBSC  is  ±  to  the  plane  GXIT.  §  518 

Let  the  plane  of  the  O  GKH  intersect  the  plane  of  the 
curve  PAQ,  in  the  straight  line  MR  ;  then  will  MR  be  J.  to 
the  plane  BBQ  {%  520),  and  therefore  ±  to  DR. 

Take  any  point  P  in  the  curve,  and  draw  SP  meeting  the 
O  G^//in  K\  join  FP,  and  draw  PM 1.  to  RM. 

Pass  a  plane  through  P  -L  to  the  axis  of  the  cone.  Let  it 
cut  the  cone  in  the  O  EPLQ,  and  the  plane  of  the  curve 
PAQ  in  the  line  PNQ. 

PJSTk  ±  to  the  plane  PSO  (§  520),  and  therefore  ±  to  DR. 

Since  PP  and  PIT  are  tangents  to  the  sphere  O,  they  are 
tangents  to  the  circle  of  the  sphere  made  by  a  plane  passing 
through  the  points  P,  P,  K,  and  are  therefore  equal.       §  246 

That  is,  PF=  PK. 

But  PJr=  LG.  %  666 

.-.  PP=  LG.  (1) 

Now  X^  and  PJtf  are  each  II  to  NR  ; 

hence  LG  is  II  to  PM.  §  485 

The  planes  GKH  ^x\^\  LPE  are  parallel.         §  491 

.-.  LG  =  PM.  §  493 

From  (1)  and  the  last  equation,  we  have 

PF=  PM. 

That  is,  any  point  P  on  the  curve  PAQ  is  equidistant  from 
a  fixed  point  Pand  a  fixed  line  RM'\xv  its  plane. 

Therefore  the  curve  PAQ  is  a  parabola. 

Q.  E.  D 


402  GEOMETRY.  —  BOOK   IX. 


Exercises. 

691.  Prove  that  if  the  abscissa  of  a  point  is  equal  to  its  ordinate,  each 
is  equal  to  the  latus  rectum. 

692.  To  draw  a  tangent  and  a  normal  at  a  given  point  of  a  parabola. 

693.  To  draw  a  tangent  to  a  parabola  parallel  to  a  given  line. 

694.  Show  that  the  tangents  at  the  ends  of  the  latus  rectum  meet 
ati). 

695.  Prove  that  the  latus  rectum  is  the  shortest  focal  chord. 

696.  The  tangent  at  any  point  meets  the  directrix  and  the  latus  rec- 
tum produced  at  points  equally  distant  from  the  focus. 

697.  The  circle  whose  diameter  is  FP  touches  the  tangent  at  A. 

698.  The  directrix  touches  the  circle  having  any  focal  chord  for 
diameter. 

699.  Given  two  points  and  the  directrix,  to  find  the  focus. 

700.  The  J.  i^  (7  bisects  TP.    (See  figure,  page  392.) 

701.  Given  the  focus  and  the  axis,  to  describe  a  parabola  which  shall 
touch  a  given  straight  line. 

702.  If  PN  is  any  normal,  and  A  PNF  is  equilateral,  then  PF  is 
equal  to  the  latus  rectum. 

703.  Given  a  parabola,  to  find  the  directrix,  axis,  and  focus. 

704.  To  find  the  locus  of  the  centre  of  a  circle  which  passes  through 
a  given  point  and  touches  a  given  straight  line. 

705.  Given  the  axis,  a  tangent,  and  the  point  of  contact,  to  find  the 
focus  and  directrix. 

706.  Given  two  points  and  the  focus,  to  find  the  directrix. 


THE   ELLIPSE.  403 


The  Ellipse. 


838.  The  locus  of  a  point  which  moves  so  that  the  sum  of 
its  distances  from  two  fixed  points  is  constant  is  called  an 
ellipse. 

The  fixed  points  are  called  the  foci,  and  the  straight  lines 
which  join  a  point  of  the  curve  to  the  foci  are  called  focal 
radii. 

The  constant  sum  is  denoted  by  2a,  and  the  distance  between 
the  foci  by  2  c.  c 

The  ratio  c :  a  is  called  the  eccentricity,  and  is  denoted  by  e. 
Therefore  c  =  ae. 

839.  Cor.  2  a  must  he  greater  than  2  c  (§  137) ;  hence  e 
must  he  less  than  1. 

840.  The  curve  may  be  described  by  the  continuous  motion 
of  a  point,  as  follows  : 


Fasten  the  ends  of  a  string,  whose  length  is  2a,  at  the  foci 
-Fand  F'.  Trace  a  curve  with  the  point  Pof  a  pencil  pressed 
against  the  string  so  as  to  keep  it  stretched.  The  curve  thus 
traced  will  be  an  ellipse  whose  foci  are  F  and  F\  and  in  which 
the  constant  sum  of  the  focal  radii  is  FF-{-  FF'. 

The  curve  is  a  closed  curve  extending  around  both  foci ;  if 
it  cuts  i^i^'  produced  in  A  and  A',  it  is  easy  to  see  that  AA^ 
equals  the  length  of  the  string. 


c 


404 


GEOMETRY.  —  BOOK   IX. 


Proposition  XI.    Problem. 

841.   To  construct  an  ellipse  hy  points,  having  given 
the  foci  and  the  constant  sum  2a, 

B 


Let  F  and  F'  be  the  foci,  and  2CD  =  2a. 

To  construct  the  ellipse. 

Oonstruction.  Through  the  foci  F,  F'  draw  a  straight  line ; 
bisect  FF*  at  0.  Lay  off  OA'  =0A  =  CD.  Then  A,  A'  are 
two  points  of  the  curve. 

Proof.   From  the  construction,  AA'=2a,  and  AF=  A'F'. 
Therefore    AF+  AF'  =  A'F-j-  A'F'=  AA'=2a, 
and  A'F-\-  A'F'=  A'F+  AF  =  AA'=  2a. 

To  locate  other  points,  mark  any  point  JT  between  F  and 
F'.  Describe  an  arc  with  F  as  centre  and  ^X  as  radius ; 
also  another  arc  with  F'  as  centre  and  A'JC  SiS  radius;  let 
these  arcs  cut  in  P,  Q. 

Then  P,  Q  are  two  points  of  the  curve. 

This  follows  at  once  from  the  construction  and  §  838. 

By  describing  the  same  arcs  with  the  foci  interchanged,  two 
more  points  P,  S  may  be  found. 

By  assuming  other  points  between  F  and  F',  and  proceed- 
ing in  the  same  way,  any  number  of  points  may  be  found. 


THE   ELLIPSE.  405 

The  curve  passing  through  all  the  points  is  an  ellipse  hav- 
ing F^  F^  for  foci,  and  2  a  for  the  constant  sum  of  focal  radii. 

aE.F. 

842.  Cor.  1.  By  describing  arcs  from  the  foci  with  the 
same  radius  OA,  we  obtain  two  points  B,  B'  of  the  curve 
such  that  they  are  equidistant  from  the  foci.  Therefore  the 
line  BB^  is  perpendicular  to  A  A*  and  passes  through  0  (§  123). 

843.  The  point  0  is  called  the  centre.  The  line  A  A*  is 
called  the  major  axis;  its  ends  A,  A^  are  called  the  vertices  of 
the  curve.  The  line  BB'  is  called  the  minor  axis.  The  length 
of  the  minor  axis  is  denoted  by  2h. 

844.  Cor.  2.  The  major  axis  is  bisected  at  0,  and  is  equal 
to  the  constant  sum  2a. 

845.  Cor.  3.   The  minor  axis  is  also  bisected  at  0  (§  123). 
Therefore  OB  =  OB'  =  b. 

846.  Cor.  4.   The  values  of  a,  b,  c  are  so  related  that 

a'  =  b'  +  c\ 
For,  in  the  rt.  A  BOF, 

bf'  =  ob'+oT. 

847.  Cor.  5.  The  axis  A  A'  bisects  FQ  at  right  angles  (§  123). 
Hence  the  ellipse  is  symmetrical  with  respect  to  its  major  axis. 

848.  The  distance  of  a  point  of  the  curve  from  the  minor 
axis  is  called  the  abscissa  of  the  point,  and  its  distance  from 
the  major  axis  is  called  the  ordinate  of  the  point. 

The  double  ordinate  through  the  focus  is  called  the  kUus 
rectwn  or  parameter. 

849.  Remark.  In  the  following  propositions  F  and  F^  denote  foci 
of  the  ellipse,  0  its  centre,  AA'  the  major  axis,  and  BB^  the  minor 
axis. 


406 


GEOMETRY.  —  BOOK   IX. 


Proposition  XIL     Theorem. 

850.  An  ellipse  is  symmetrical  with  respect  to  its 
minor  axis. 


A'        F'  0 

Let  P  be  a  point  of  the  curve,  PDQ  be  perpendicular 
to  OB,  meeting  OB  in  D,  and  let  DQ  equal  DP. 

To  prove  that  Q  is  also  a  point  of  the  curve. 

Proof.  Join  P  and  Q  to  the  foci  F,  F'. 

Eevolve  ODQF  ahout  OB  ;  Pwill  fall  on  P'  and  Q  on  P. 

Therefore  Q,F=  PF\ 

and  Z  PQF=  Z  QPF'. 

Therefore  A  PQF  =■  A  QPF\ 

and  QF'  =  PF. 

Hence  QF-\-  QF'  =  PF-\-  PF. 

But  PF-\-  PF  =  2a. 

Therefore  QF-{-  QF'  =  2a. 

Therefore  Q  is  a  point  of  the  curve. 

a  E.  D. 

851.   Every  chord  passing  through  the  centre  of  an  ellipse 

is  called  a  diameter. 

862.  Cor.  1.  From  §§  847,  850,  it  follows  that  an  ellipse 
consists  of  four  equal  quadrantal  arcs  symmetrically  placed 
about  its  centre  0  (§§  209,  64). 

853.   Cor.  2.   Every  diameter  is  bisected  at  the  centre. 


§106 


Hyp. 


THE   ELLIPSE. 


407 


Proposition  XIIL     Theorem. 

854.  Ifd  denotes  the  abscissa  of  a  point  of  an  ellipse, 
r  and  r'  its  focal  radii,  then  r'  =  a  +  ed,  r=a  —  ed. 


Let  P  be  any  point  of  an  ellipse,  PM  perpendicular  to 
AA\  d  equal  OM,  r  equal  PF,  r'  equal  PF*. 

To  prove  r'  =  a  -\-  ed,     r  =  a  —  ed. 

Proof.   From  the  rt.  A  i^PJf  and  F'PM 

Therefore         r''^  -  r'  =  PW  -  FM\ 

Or       (r'  +  r)  (r'  -r)  =  (F'M-^  F3I)  (F'3I-  FM). 


Now 
Also, 
Hence 

From 


/  +  r  =  2a,  and  F^M-\-  FM=  2c. 
F'M-  FM=  OF'-i-OM-  FM=20M=  2d 
a  (r'  —  r)  =  2  cd. 


t            2cd     ci    ■, 
r  —  r  = =  2ed. 


r'  -f  r  =  2  a,  and 


2ed, 


2r'=2(a  +  ec?),  and  2r  =  2{a~ed). 
Therefore         r'  =  a  -f  ec?,  and  r  =  a  —  ed. 


408 


GEOMETRY.  —  BOOK    IX, 


855.  The  circle  described  upon  the  major  axis  of  an  ellipse 
as  a  diameter  is  called  the  auxiliary  circle.  The  points  where 
a  line  perpendicular  to  the  major  axis  meets  the  ellipse  and  its 
auxiliary  circle  are  called  corresponding  points. 


Proposition  XIV.     Theorem. 

856.  The  ordinates  of  two  corresponding  points  in 
an  ellipse  and  its  auxiliary  circle  are  in  the  ratio 
h:a. 


Let  P  be  a  point  of  the  ellipse,  Q  the  corresponding 
point  of  the  auxiliary  circle,  and  QP  meet  AA'  at  M, 


To  prove 
Proof.   Let 
then 


PM:  QM^h'.a. 
OM^d) 
'  d\ 


QM  -  a^ 

PM'  =  PF'-  FM"  =  {a-  edy-{c -  dj     §  854 
=  a'  —  '2.aed-\-e^d''  —  c^ -\-  2cd  —  d\ 
Or,  since  c  =  ae  and  c^  —  &  =  J',  §  846 

PM'^h^-{\-e')d^^^(a^-d-'\ 
a^ 

Therefore  PW :  QW  =  b' :  a\ 

Or  PM  .QM  ^b  :a, 

aE.  D. 


THE    ELLIPSE. 


409 


Proposition  XV.     Problem. 

857.  To  construct  an  ellipse  hy  points,  having  given 
its  two  axes. 


A'  O  M  A 

Let  OA,  OB  be  the  given  semi- axes,  0  the  centre. 

Oonstniction.  With  0  as  centre,  and  OA,  OB,  respectively, 
as  radii,  describe  circles. 

From  0  draw  any  straight  line  meeting  the  larger  circle  at 
Q  and  the  smaller  circle  at  R, 

Through  Q  draw  a  line  II  to  OB,  and  through  B  draw  a 
line  II  to  OA. 

Let  these  lines  meet  at  P. 

Then  will  P  be  a  point  of  the  required  ellipse. 

Proof.   If  QP  meet  A  A*  at  M, 

PM:  QM=  OB  :  OQ.  §  309 

But  OB  =  b  and  OQ  =  a. 

Therefore  PM:  QM=  h  :  a. 

Therefore  P  is  a  point  of  the  ellipse.  (§  856) 

By  drawing  other  lines  through  0,  any  number  of  points  on 
the  ellipse  may  be  found  ;  a  smooth  curve  drawn  through  all 
the  points  will  be  the  ellipse  required.  o.  e.  f. 


410 


GEOMETRY. 


BOOK    IX. 


Proposition  XVI.     Theorem. 

858.  The  square  of  the  ordinate  of  a  point  in  an 
ellipse  is  to  the  product  of  the  segments  of  the  major 
axis  made  hy  the  ordinate  as  IP- 1  c^. 


Let  P,  Q  be  corresponding^  points  in  the  ellipse  and 
auxiliary  circle,  respectively;  let  PQ  meet  AA!  in  M, 

To  prove  ¥W  :  ^Jf  X  A'M^  b"  :  a\ 

Proof.  FM''  :  QM"  =  h'  :  a\  §  856 

But  03"  =  AMY.  A'M.  §  337 

Therefore  PM"  :  AMv.  A' 31=  h'  :  a\  q.  e.  d. 

859.   Cor.   The  latus  rectum  is  a  third  proportional  to  the 
major  axis  and  the  minor  axis. 


For 

Now 
and 

Therefore 

Hence 
and 

Therefore 


LF-.AFxA'F=b':a\ 
A'F=  a-c, 
AF=a  +  c. 
AFxA^F=a'~c'  =  b\ 
LT  :h''  =  h':  a\ 
LF:h  ^h   :a. 
2a'.2h=r2h:2LF. 


§858 


§846 


THE   ELLIPSE. 


411 


Proposition  XVII.     Theorem. 

860.  The  sum  of  the  distanees  of  any  point  from 
the  foei  of  an  ellipse  is  greater  or  less  than  2  a,  oa^- 
cording  as  the  point  is  without  or  within  the  curve. 

Q 


A'        F' 

1.  Let  Q  be  a  point  without  the  curve. 

To  prove  QF+  QF'  >  2a. 

Proof.   Let  P  be  any  point  on  the  arc  of  the  curve  between 
QF and  QF'.     Draw  PPand  FF'. 
Then  QF+  QF'  >  FF+  FF'.  §  118 

But  FF-{-FF=2a.  §838 

Therefore  QF-\-  QF'  >  2  a. 

2.  Let  Q'  be  a  point  within  the  curve. 

To  prove  Q'F-{-  Q'F'  <  2  a. 

Proof.   Let  F  be  any  point  of  the  curve  between  FQ^  and 
F^Q'  produced. 
Then  Q'F-h  Q'F'  <  FF+  FF'.  §  118 

But  PP+PP'=2a. 

Therefore  Q'F+  Q'F'<  2  a.  o.  e,  d. 

861.  Cor.  Converse^,  a  point  is  without  or  within  an  ellipse 
according  as  the  sum  of  its  distances  from  the  foci  is  greater  or 
less  than  2  a. 

862.  A  straight  line  which  touches  but  does  not  cut  an 
ellipse  is  called  a  tangent  to  the  ellipse.  The  point  where  it 
touches  the  ellipse  is  called  the  point  of  contact. 


412 


GEOMETRY.  — •  BOOK   IX. 


Proposition  XVIII.    Theorem. 

If  through  a  -point  P  of  an  ellipse  a  line  he 
drawn  bisecting-  the  angle  between  one  of  the  focal 
radii  and  the  other  produced,  every  point  in  this 
line  except  P  is  without  the  curve^ 


Let  FT  bisect  the  angle  PPG  between  FT  and  FP 
produced,  and  let  Q  be  any  point  in  PT  except  P, 

To  prove  that  Q  is  without  the  curve. 

Proof.        Upon  FP  produced  take  PG  =  PF'. 

Join  OF',  QF,  QF',  QG. 

Then  QG+QF>GF'. 

Now  AGPQ  =  AF'PQ. 

Therefore  QG  =  QF'. 

Also  GF=2a. 

Therefore  QF'  +QF>2a. 

Therefore  Q  is  without  the  curve. 


§137 
§150 

§638 

§861 
aE.  D. 

864.  Cor.  1.  PTis  the  tangent  at  P.  §  862 

865.  Cor.  2.   The  tangent  to  an  ellipse  at  any  point  bisects 
the  angle  between  one  focal  radius  and  the  other  produced. 


THE    ELLIPSE.  ,  413 

866.  Cor.  3.  If  QF'  cuts  PT  at  X,  then  GX-=  F'X,  and 
FT  is  perpendicular  to  QFK  §  123 

867.  Cor.  4.   The  locus  of  the  foot  of  a  perpendicular  dropped 
from  the  focus  of  an  ellipse  to  a  tangent  is  the  auxiliary  circle. 

For,  join  OX. 

Since  F'X=  GX,  and  F'0=OF, 

therefore  0X=  \FG=  i(2a)  =  a.  §  189 

Therefore  the  point  X  lies  in  the  auxiliary  circle. 

'     Proposition  XIX.     Problem. 

868.  To  draw  a  tangent  to  an  ellipse  from  an  exte^ 
rior  point.  -^.^ 


To  draw  tangents  to  the  ellipse  ORQfroTn  the  exterior  point  P. 

Oonstruction.  Describe  arcs  with  P  as  centre  and  PF  as 
radius,  and  with  F^  as  centre  and  2a  as  radius;  let  these 
arcs  intersect  in  G  and  8. 

Join  OF^  and  8F',  cutting  the  curve  in  Q  and  R  respec- 
tively. 

Join  QP  and  RP,  and  they  will  be  the  tangents  required. 

Proof.  PG  =  PF,  and  QG  =  QF.         Cons.,  §  838 

.•.APQG  =  APQF.  §160 

.■.ZPQG  =  APQF. 

Therefore  PQ  is  the  tangent  at  Q. 

For  like  reason   Pi2  is  the  tangent  at  i?.  aE.F. 

869.  Cor.  The  (D  GFS  and  GS  will  always  intersect 
(Ex.  78).  Hence,  two  tangents  may  always  be  drawn  to  an 
ellipse  from  an  exterior  point. 


414 


GEOMETRY. 


BOOK   IX. 


Proposition  XX.     Theorem. 

870.  The  tangents  drawn  at  two  corresponding 
points  of  an  ellipse  and  its  auxiliary  circle  cut  the 
7}vajor  axis  produced  at  the  same  point. 


Af       F' 


0     N  D       M    F   A 


Let  the  tangent  to  the  auxiliary  circle  at  Q  cut  the 
major  axis  produced  at  T,  and  let  the  ordinate  QM 
meet  the  ellipse  at  P.    Draw  PT. 

To  prove  that  FT  is  the  tangent  to  the  ellipse  at  P. 

Proof.  Through  R,  any  point  in  PT  except  P,  draw  RD  A. 
to  AA\  cutting  the  tangent  QT,  the  auxiliary  circle,  and  the 
ellipse,  in  L,  K,  and  xS',  respectively. 

Then         RD  :  FM=  DT:  MT^ 

or  RB'.LD^PM 

But  FM:QM-=h 

.-.  RD.  LD  =  b 

Again,  8D:KD  =  b 

.'.  RD:  LD  =  SD: 

But  DD  >  KD. 

:.  RD  >  SD. 

.'.  R  is  without  the  ellipse. 

Hence  FT  is  the  tangent  at  F.  §  862 

aE.  D 


=  LD  :  QM, 

§321 

QM. 

§208 

a. 

§856 

a. 

a. 

§856 

KD. 

THE   ELLIPSE. 

871.   Cor.  1.  OTxOM=a\ 


415 
§334 


872.  The  straight  line  PN  drawn  through  the  point  of  con- 
tact of  a  tangent,  perpendicular  to  the  tangent,  is  called  the 
normal. 

MT  ia  called  the  subtangeni,  MJ^  the  subnormal. 

873.  Cor.  2.  The  normal  bisects  the  angle  between  the  focal 
radii  of  the  point  of  contact. 

For  Z  TPN=  Z  GPN=.  90°.  §  872 

Subtract  Z  TPF--=  A  GPF\  %  865 

And  Z  FPN=  Z  F'PN. 

Hence  a  ray  of  light  issuing  from  F  will  be  reflected  to  F\ 

874.  Cor.  2f.  If  d  denote  the  abscissa  of  the  point  of  contact, 
the  distances  measured  on  the  major  axis  from  the  centre  to  the 

tangent  and  the  normal  are  —  and  id,  respectively. 

§871 


§334 


(1)  Since 

OM=d,  and  OTx  OM=d 

therefore 

d 

(2)  Since 

OMx  MT=  QM\ 

and 

MNx  MT=  PM\ 

therefore 

OM       Qir     a' 

MN     PM'     b^ 

Therefore 

OM~MN     a'-h"     e      ^ 
OM               a'         a"        ' 

That  is, 

OM       ' 

Hence 

ON^e'xOM=--e'd. 

§301 


416 


GEOMETRY.  —  BOOK    IX. 


Proposition  XXI.     Problem. 

875.   The  tangents  drawn  at  the  ends  of  any  diam- 
eter are  parallel  to  eaeh  other. 


Let  POQ  be  any  diameter,  PT  and  QT  the  tangents 
at  P,  Q  respectively,  meeting  the  major  axis  at  T,  V, 


To  'prove 

priito  qv. 

Proof. 

Draw  the  ordinates  PM,  QN. 

.  Then 

A  OPM=  A  OQJSr. 

§148 

Therefore 

0M=  ON. 

But 

^^=oV^"^^^'=oV 

§874 

Hence 

OT=OT\ 

Therefore 

AOPT=AOQT\ 

§150 

and 

AOPT^Aoqr, 

Hence 

PT\%  11  to  qr. 

aE.D. 

876.   One  diameter  is  conjugate  to  another,  if  the  first  is 
parallel  to  the  tangents  at  the  extremities  of  the  second. 
Thus  if  ROB\^  I!  to  PT,  PS  is  conjugate  to  Pq. 


THE   ELLIPSE. 


417 


Proposition  XXII.     Theorem. 

877.   If  one  diameter  is  conjugate  to  a  second,  the 
second  is  conjugate  to  tJie  first. 


Let  the  diameter  POP'  be  parallel  to  the  tangent  RT. 
To  prove  that  MOJR'  is  parallel  to  the  tangent  PT. 

Proof.   Draw  the  ordinates  PM  and  RN,  and  produce  tliem 
to  meet  the  auxiliary  circle  in  Q  and  8. 

Join  OP,  OQ,  OB,  OS;  and  draw  the  tangents  QT,  ST'. 
Now,  since  OP  is  II  to  i?X'. 

the  A  03fP  and  T'JSTP  are  similar.  §  321 

.-.  T'JST:  OM=NE:  MP. 
But.  NR  :  NS  =  MP :  MQ,  §870 

or  NR  :  MP=  NS  :  MQ. 

.-.  T'J^:  OM^NS  :MQ. 
Hence  A  T'JSTS  and  OMQ  are  similar.  §  326 

.  .'.Z]^T'S=ZMOQ. 

.'.  T'Sk  II  to  OQ.  §105 

Hence  Z  QOS  =  Z  OST'  =  90°.  §  240 

.-.  S0\^  W  io  QT  §  105 


418 


GEOMETRY.  —  BOOK    IX. 

/.A  SNO  and  QMT&re  similar. 
,\0N:  TM=NS  :MQ, 

^JSrH'.MR  §856 

.*.  A  ONR  and  IMP  are  similar. 
.-.  OR  is  a  to  FT. 
.'.  RR^  is  conjugate  to  PP*.  aE.D. 


878.  Cor.  1.  Angle  QOS  is  a  right  angle. 

879.  Cor.  2.   MP  :  0N=  b  :  a. 


For 

08=0Q, 

and 

OS  is  ±  to  OQ, 

§878 

Also 

OM  is  ±  to  NS. 

.\ZNSO  =  ZMOQ. 

§  113,  Rem. 

Hence 

A  JSrSO  =  A  MOQ, 
.'.  0N=  MQ. 
.'.  MP  :  0]Sr=  MP  :  MQ. 

§148 

But 

MP:  MQ  =  b:a. 

§856 

Hence 

MP :  ON^h:  a. 

THE    ELLIPSE. 


419 


Proposition  XXIII.     Theorem. 
The  area  of  an  ellipse  is  equal  to  Trob, 


A'  M        N  0 

Let  A'PRA  he  any  semi- ellipse. 

To  prove  that  the  area  of  twice  A'PRA  is  equal  to  irab. 
Proof.    Let  FM,  i?iVbe  two  ordinates  of  the  ellipse,  and  let 
Q,  S  be  the  corresponding  points  on  the  auxiliary  circle. 
Draw  PV,  QUW  to  the  major  axis,  meeting  iViS'in  F,  U. 
Then  O  PiV=  FMx  MN, 

and  nQN=-QMxMN. 

O  PN      PMxMN     PM     h 

a 


Therefore 


§856 


CJ  Q]^      QMxMN     QM 
The  same  relation  will  be  true  for  all  the  rectangles  that 
can  be  similarly  inscribed  in  the  ellipse  and  auxiliary  circle. 

Hence  sum  of  g?  ip  elHpse^ft,  g  303 

sum  of  £17  in  circle       a 
And  this  is  true  whatever  be  the  number  of  the  rectangles. 
But  the  limit  of  the  sum  of  the  (H  in  the  ellipse  is  the  area 
of  the  ellipse,  and  the  limit  of  those  in  the  0  is  the  area  of 
the  O. 

Therefore  area  of  ellipse  ^  h  g  260 

area  of  circle       a 


Therefore  the  area  of  the  ellipse  —  -  X  tto^ 

a 


irab.      §  425 
aE.  o. 


420 


GEOMETRY.  —  BOOK   IX. 


Proposition  XXIV.     Theorem. 

^l)  The  section  of  a  right  circular  cone  made  hy  a 
plane  that  cuts  all  the  elements  of  the  surface  of  the 
cone  is  an  ellipse. 


Let  APA*  be  the  curve  traced  on  the  surface  of  the 
cone  SBC  by  a  plane  that  cuts  all  the  elements  of 
the  surface  of  the  cone. 

To  prove  that  the  curve  A  PA*  is  an  ellipse. 

Proof.  The  plane  passed  through  the  axis  of  the  cone  and 
X  to  the  secant  plane  APA*  cuts  the  surface  of  the  cone  in 
the  elements  8B,  SO,  and  the  secant  plane  in  the  line  AA'. 

Describe  the  ©  0  and  O*  tangent  to  SB,  SO,  AA\  Let  the 
points  of  contact  be  D,  II,  F,  and  B,  C,  F\  respectively. 

Turn  BSC  and  the  ©  0,  0'  about  the  axis  of  the  cone. 
The  lines  SB,  SO  will  generate  the  surface  of  a  right  circular 
cone  cut  by  the  secant  plane  in  the  curve  APA' ;  and  the 
©  0,  O  will  generate  spheres  which  touch  the  cone  in  the 
©  DNH,  BWQ,  and  the  secant  plane  in  the  points  F,  F'^ 


THE   ELLIPSE.  421 

Let  P  be  any  point  on  the  curve  APA\  Draw  PF,  PF'; 
and  draw  jSP,  which  touches  the  (D  PIT,  PC  at  the  points  iV^, 
iV',  respectively. 

Since  PF  and  PiV^  are  tangent  to  the  sphere  0,  they  are 
tangent  to  the  circle  of  the  sphere  made  by  a  plane  passing 
through  P,  F,  and  iV. 

Therefore  PF=  PK  §  246 

Likewise  PF'  =  PJV'. 

Hence  PP+  PF'  =  PN^-  PJST' 

=  NN',  a  constant  quantity. 

Therefore  APA'  is  an  ellipse  with  the  points  F  and  F'  for 
foci,  and  A  A'  as  2  a.  q,  e,  q, 

882,  Cor.  If  the  secant  plane  is  parallel  to  the  base,  the  sec- 
tion is  a  circle,  which  is  a  particular  case  of  the  ellipse. 


Exercises. 


707.  Prove  that  the  major  axis  is  the  longest  chord  that  can  be  drawn 
in  an  ellipse. 

708.  If  the  angle  FBF^  is  a  right  angle,  prove  that  a?  =  262. 

709.  To  draw  a  tangent  and  a  normal  at  a  given  point  of  an  ellipse. 

710.  To  draw  a  tangent  to  an  ellipse  parallel  to  a  given  straight  line. 

711.  Given  the  foci ;  it  is  required  to  describe  an  ellipse  touching  a 
given  straight  line. 

712.  Prove  that  OF"^  =OTxON.    (See  figure,  page  414.) 

713.  Prove  that  OM :  0N=  a^ :  c^.     (See  figure,  page  414.) 

714.  The  minor  axis  is  the  shortest  diameter  of  an  ellipse. 

715.  At  what  points  of  an  ellipse  will  the  normal  at  the  point  pass 
through  the  centre  of  the  ellipse  ? 

716.  Prove  that  if  FR,  F^S  are  the  perpendiculars  dropped  from  the 
foci  to  any  tangent,  then  FR  X  F^S-=  b'^. 


422  GEOMETRY.  —  BOOK    IX. 

717.  To  draw  a  diameter  conjugate  to  a  given  diameter  in  a  given 


718.  Given  2a,  2b,  one  focus,  and  one  point  of  the  curve,  to  construct 
the  curve, 

719.  If  from  a  point  Pa  pair  of  tangents  FQ  and  FE  be  drawn  to  an 
ellipse,  then  FQ  and  FF  will  subtend  equal  angles  at  either  focus. 

720.  To  find  the  foci  of  an  ellipse,  having  given  the  major  axis  and 
one  point  on  the  curve. 

721.  To  find  the  foci  of  an  ellipse,  having  given  the  major  axis  and 
a  straight  line  which  touches  the  curve. 

722.  A  straight  line  moves  so  that  its  extremities  are  always  in  con- 
tact with  two  fixed  straight  lines  perpendicular  to  each  other.  Prove 
that  any  point  of  the  moving  line  describes  an  ellipse. 

723.  To  construct  an  ellipse,  having  given  one  of  the  foci  and  three 
tangents. 

724.  To  construct  an  ellipse,  having  given  one  focus,  two  tangents, 
and  one  of  the  points  of  contact. 

725.  To  construct  an  ellipse,  having  given  one  focus,  one  vertex,  and 
one  tangent. 

726.  The  area  of  the  parallelogram  formed  by  drawing  tangents  to 
an  ellipse  at  the  extremities  of  any  pair  of  conjugate  diameters  is  equal 
to  the  rectangle  contained  by  the  axes  of  the  ellipse. 


THE    HYPERBOLA.  423 


The  HyrERBOLA. 

883.  The  locus  of  a  point  which  moves  so  that  the  difference 
of  its  distances  from  two  fixed  points  is  constant  is  called  an 
hyperbola. 

The  fixed  points  are  called  the  foci,  and  the  straight  lines 
which  join  a  point  of  the  locus  to  the  foci  are  called  focal 
radii. 

The  constant  difference  is  denoted  by  2  a,  and  the  distance 
between  the  foci  by  2c. 

The  ratio  c :  a  is  called  the  eccentricity,  and  is  denoted  by  e. 
Therefore  c  =  ae. 

884.  Cor.  2a  micst  be  less  than  2c  (§  137) ;  hence  c  7)iust  be 
greater  than  1. 

885.  An  hyperbola  may  be  described  by  the  continuous 
motion  of  a  point,  as  follows  : 


To  one  of  the  foci  F'  fasten  one  end  of  a  rigid  bar  F'B  so 
that  it  is  capable  of  turning  freely  about  F*  as  a  centre  in  the 
plane  of  the  paper. 


424  GEOMETEY.  —  BOOK   IX. 

Take  a  string  whose  length  is  less  than  that  of  the  bar  by 
the  constant  difference  2a,  and  fasten  one  end  of  it  at  the  other 
focus  F^  and  the  other  end  at  the  extremity  B  of  the  bar. 

If  now  the  rod  is  made  to  revolve  about  F^  while  the  string 
is  kept  constantly  stretched  by  the  point  of  a  pencil  at  P,  in 
contact  with  the  bar,  the  point  P  will  trace  an  hyperbola. 


For,  as  the  bar  revolves,  F^P  and  FP  are  each  increas- 
ing by  the  same  amount ;  namely,  the  length  of  that  portion 
of  the  string  which  is  removed  from  the  bar  between  any  two 
positions  of  P\  hence  the  diiference  between  F^P  and  FP 
will  remain  constantly  the  same. 

The  curve  obtained  by  turning  the  bar  about  F  is  the  right- 
hand  branch  of  the  .hyperbola.  Another  similar  branch  on 
the  left  may  be  described  in  the  same  manner  by  making  the 
bar  revolve  about  Pas  a  centre. 

If  the  two  branches  of  the  hyperbola  cut  the  line  FF'  at  A 
and  A\  it  is  easy  to  see,  from  the  symmetry  of  the  construction, 
that  ^^'=  2  a. 

The  hyperbola,  therefore,  is  not  a  closed  curve,  like  the 
ellipse,  but  consists  of  two  similar  branches  which  are  sepa- 
rated at  their  nearest  points  by  the  distance  2a,  and  which 
recede  indefinitely  from  the  line  PP'  and  from  one  another. 


THE   HYPERBOLA. 


425 


Proposition  XXV.     Problem. 

886.   To  construct  an  hyperbola  hy  -points,  having 
given  the  foci  and  the  constant  difference  2a. 


^\Xi/' 

\^ 

/ 

\ 
/            \ 

.1  M 

> 

\ 

1  rrX 

\      \    1 

1'  0 

A 

j^ 

— -4^^ 

-"K" 

/s\ 

/qn 

\ 

Let  F,  F'  be  the  foci,  and  a  =■  CD. 

To  construct  the  hyperbola. 

Construction.       Lay  off  OA  =  0A'=  CD. 

Then  A  and  A^  are  two  points  of  the  curve. 

For  from  the  construction  AA^  =  2a  and  AF—  A'F'. 

Therefore    AF'  -  AF=  AF*  -  A'F'  =  AA'  =  2a. 

And  A^F  -  A'F' =  A'F-  AF=  AA'  =  2a. 

To  locate  other  points,  mark  any  point  JTin  i^'i^  produced. 
Describe  arcs  with  F'  and  i^as  centres,  and  A'X  and  ^Xas 
radii,  intersecting  in  P,  Q. 

Then  P,  Q  are  points  of  the  curve. 

By  describing  the  same  arcs  with  the  foci  interchanged,  two 
more  points  P,  8  may  be  found. 

By  assuming  other  points  in  P'P  produced,  any  number  of 
points  may  be  found. 

The  curve  passing  through  all  the  points  thus  determined 
is  an  hyperbola  having  FF'  for  foci  and  2  a  for  the  constant 
difference  of  the  focal  radii.  o.  e.  f, 


426 


GEOMETRY.  —  BOOK    IX. 


887»  Cor.  1.  No  point  of  the  curve  can  be  situated  on  the 
perpendicular  to  FF^  erected  at  0.  For  every  point  of  this 
perpendicular  is  equidistant  from  the  foci. 

888.  The  point  0  is  called  the  centre ;  A  A'  is  called  the 
transverse  axis;  A  and  A^  are  called  the  vertices. 

In  the  perpendicular  to  FF^  erected  at  0,  let  B^  B^  be  the 
two  points  whose  distance  from  A  (or  J.')  is  equal  to  c ;  then 
BB*  is  called  the  conjugate  axis,  and  the  length  BB'  is 
denoted  by  2b. 

If  the  transverse  and  conjugate  axes  are  equal,  the  hyper- 
bola is  said  to  be  equilateral  or  rectangular. 


a 


889.  Cor.  2.  Both  the  axes  are  bisected  at  0. 

890.  Cor.  3.  It  is  evident  thai  c^  —  a^-\-  IP. 

891.  Cor.  4.  The  curve  is  symmetrical  ivith  respect  to  the 
transverse  axis.. 

892.  The  distances  of  a  point  of  the  curve  from  the  trans- 
verse and  conjugate  axes  are  called  respectively  the  ordinate 
and  abscissa  of  the  point.  The  double  ordinate  through  the 
focus  is  called  the  latus  rectum  or  parameter. 


893.    Remark.  The  letters  A,  A\  B,  5',  F,  F\  and  0,  will  be  used 
to  designate  the  same  points  as  in  the  above  figure. 


THE   HYPERBOLA. 


427 


Proposition  XXVI.     Theorem. 

894.  An  hyperbola  is  synvinetrical  with  respect  to 
its  conjugate  axis. 


F'  A 


Let  P  be  a  point  of  the  curve,  PDQ  be  perpendicu- 
lar to  OB,  meeting  OB  at  D,  and  let  DQ  equal  DP. 

To  prove  thai  Q  is  also  a  point  of  the  curve. 

Proof.  Join  P  and  Q  to  the  foci  F,  F'. 

Turn  ODQF'  about  OB ;  F'  will  fall  on  F,  and  Q  on  P. 

Therefore  QF'  =  FF, 

and  ZFQF'  =  Z.QFF. 

Therefore  A  FQF'  =  A  QFF, 

and  QF=  FF'. 

Hence  QF-  QF'  =  FF'  -  FF. 

But  FF'-FF=2a. 

Therefore  QF-  QF'  =  2a. 

Therefore  Q  is  a  point  of  the  curve. 

895.  Every  chord  passing  through  the  centre  is  called  a 
diameter. 

896.  Cor.  1.  An  hyperbola  consists  of  four  equal  quadrantal 
arcs  symmetrically  placed  about  its  centre  0.  §  209 

897.  Cor.  2.  Every  diameter  is  bisected  at  0. 


§150 


Hyp. 


Q.  E.  O. 


428 


GEOMETRY.  —  BOOK   IX. 


Proposition  XXVII.     Theorem. 

898.  If  d  denote  the  abscissa  of  a  -point  of  an  hy- 
perbola,  r  and  r'  its  focal  radii,  then  r^ed  —  a,  and 
r'=  ed+  a. 


A! 


0 


A     F 


M 


Let  P  be  any  point  at  the  curve,  FM  perpendicular 
to  AA\  d  equal  OM,  r  equal  PF,  r'  equal  PF', 

To  prove  r  =  ed—a,     r'  =  ed-\-a. 

Proof.   From  the  rt.  A  FFM,  F'FM, 


F]IF  +  FM\ 


FM'  +  F'IF 

2 


Therefore  r'^  -r'  =  F'M   -  FM 

Or  (r'  +  r)  (r'  -r)  =  (F'M-{-  FM)  {F'M-  FM). 

Now  r'-r  =  2a,  and  F'M~FM^2c. 

Also  F'M+  FM=  2  0F-\-  2  FM=  2  0M=  2d. 

By  substituting  these  values, 

a  (r'  +  r)  =  2  cd. 

Or  r'  +  r  =  — =  2cc^. 

a 

From  r'+ r  =  2ec?,  and /  — 7'=2a, 
by  addition,  2  r'  =  2  {ed-\-  a) ; 

by  subtraction,  2r  =2(ed—a). 

Therefore  r  =  ed—a,  and  r*  =  ed-i-a.  o. e. o. 


THE    HYPERBOLA„ 


429 


899,   The  circle  described  upon  AA'  as  a  diameter  is  called 
the  auxiliary  circle. 


Proposition  XXVIII.    Theorem. 

900.  Any  ordinate  of  an  hyperbola  is  to  the  tangent 
from  its  foot  to  the  auxiliary  circle  as  h  is  to  a. 

v/ 


Let  P  be  any  point  of  the  hyperbola,  PM  the  drdi' 
nate,  MQ  the  tangent  drawn  from  M  to  the  auxiliary 
circle. 

FM:  QM^h'.a. 

OM^d. 

QW  =  d'-  a\ 


To  prove 
Proof.   Let 
Then 
Also 


Or  since 

Therefore 
Or 


=  (ed-ay-{d-cy  §898 

=  e^d'-2aed-\-d'-d'-{-2cd-c\ 
c  =  ae,  and  a^-c'=-&^  §  890 

FM'=(e''-l)d'-b'=^(d'-a:'). 


FM'  :  QM'  =  b'  :  a\ 
FM  .QM=^h   '.a. 


aE.t>. 


430 


GEOMETRY. 


BOOK    IX. 


Proposition  XXIX.     Theorem. 

901.  The  square  of  the  ordinate  of  a  point  in  an 
hyperbola  is  to  the  product  of  the  distances  from  the 
foot  of  the  ordinate  to  the  vertices  as  h^  is  to  a^- 

P/ 


Let  P  be  any  point  of  the  curve,  PM  the  ordinate, 
MQ  the  tangent  drawn  from  M  to  the  auxiliary  circle. 

To  prove  PM'  :  AMx  A'M=  b'  :  d\ 
Proof.   Now  Fif  :  QM'  =  b'  :  a\ 

But  QW  =  AMx  A'M. 

Therefore  FM'  :  AMx  A'M=  b'  :  a\ 


§348 


902.    Cor.   The  laius  rectum  is  a  third  proportional  to  the 
transverse  and  conjugate  axes. 

For  LF^  :  AF X  A'F=  b'  :  a\ 

But  AF=  c  —  a,  and  AF'=  c  +  a. 


Therefore  AFx  A'F=  &  -  c 

Hence  LT  \W  =  l^\  a\ 

And  LF\b  =^h  '.a. 

Therefore  2a  :  26  =  25  :  2  LF. 


b\ 


§901 
§890 


THE    HYPERBOLA. 


431 


Proposition  XXX.     Theorem. 

903.  The  difference  of  the  distances  of  any  point 
froT)%  the  foci  of  an  hyperbola  is  greater  or  less  than 
2a,  according  as  the  point  is  on  tJie  concave  or  con- 
vex side  of  the  curve. 


F'  A  o  ^L    F 

1.  Let  Qbe  a  point  on  the  concave  side  of  the  curve. 
To  prove  QF'  —QF>2a. 

Proof.  Let  QF'  meet  the  curve  at  P. 

Since     F'Q=^F'F+FQ,  and  FQ<FF-j-FQ, 
therefore  F'Q  -  -  FQ  >  F'F  -  FF. 

But  F'F-FF=-2a. 

Therefore  F'Q-FQ>2  a. 

2.  Let  Q'  be  a  point  on  the  convex  side  of  the  curve 

To  j^rove  Q'F'  —  Q'F<  2  a. 

Proof.  Let  Q'F  cut  the  curve  at  F. 

Since   F'Q'  <  F'Fi-FQ',  and  FQ'  -  PP+  FQ', 
therefore  F'Q' -  FQ' <  F'F  -  FF. 

But  F'F-FF=2a. 

Therefore  F'Q'~FQ'<2a.  o.  e.  d. 

904.  Cor,  Convei^sely ,  a  point  is  on  the  concave  or  the  con- 
vex side  of  the  hypei^hola  according  as  the  differejice  of  its  dis- 
tances from  the  foci  is  greater  or  less  than  2  a. 


432 


GEOMETRY. 


BOOK    IX. 


905.  A  straight  line  which  touches  but  does  not  cut  the 
Jiyperbola  is  called  a  tangent,  and  the  point  where  it  touches 
the  hyperbola  is  called  the  point  of  contact. 


Proposition  XXXL    Theorem. 

906.  If  through  a  point  P  of  an  hyperbola  a  line  he 
drawn  bisecting  the  angle  between  the  focal  radii^ 
every  point  in  this  line  except  P  is  on  the  convex  side 
of  the  curve. 


F'  A 


Let  PT  bisect  the  an^le  FPF',  and  let  Q  be  any  point 
in  PT  except  P. 

To  prove  that  Q  is  on  the  convex  side  of  the  curve. 

Proof.   On  PF'  take  PG  =  PF;  draw  FG,  QF,  QF\  QG. 


Then 

QF'  ~QG<  GF\ 

Also 

APGQ  =  APFQ. 

Therefore 

QG=QF. 

Also 

GF'=PF'-PF=2a 

Therefore 

QF'-QF<2a. 

§137 
§150 


§904 
ae.  D. 

§905 

908.  Cor.  2.   The  tangent  to  an  hyperbola  at  any  point  bisects 
the  angle  between  the  focal  radii. 

909.  Cor.  3.   The  tangent  at  A  is  perpendicular  to  A  A*. 

910.  Cor.  4.  If  FG  cuts  PT  at  X,  then  GX^  FX,  and 
PT  IS  perpendicular  to  FG^ 


Therefore  Q  is  on  the  convex  side  of  the  curve. 
907.   Cor.  1.  PTis  the  tangent  at  P. 


THE    HYPERBOLA. 


433 


911.    Cor.  5.    The  locus  of  the  foot  of  the  perpendicular  from 
the  focus  of  an  hyperbola  to  a  tangent  is  the  auxiliary  circle. 

For,  since         FX=  GX,  and  FO  =  OF', 
therefore        0X=  iF'G  =  i  (FF'  -  FF)  -  a.  §  189 

Therefore  the  point  X  lies  in  the  auxiliary  circle. 


Proposition  XXXII.     Problem. 

912.    To  draw  a  tangent  to  an  hyperbola  from  a 
§iven  exterior  point. 


Let  P  he  the  given  point. 

Oonstrnction.  Describe  arcs  with  F  as  centre  and  FF  as 
radius,  and  with  F'  as  centre  and  2  a  as  radius ;  let  these  arcs 
intersect  in  O  and  F. 

Draw  F'O  and  F'F,  and  produce  them  to  meet  the  curve 
in  Q  and  D,  respectively. 

Join  FQ  and  FD ;  FQ  and  FD  are  the  tangents  required. 
Proof.        FG  =  FF,  QF=  QF' -2a=^- QG. 
.\AFQG--=AFQF. 
.-.  Z  FQG  -  Z  FQF. 
.'.  FQ  is  the  tangent  at  Q. 
For  like  reason  FD  is  the  tangent  at  D.  o.  e.  f 

913.  Cor.  Two  tangents  may  always  he  drawn  to  an  hyper- 
bola from  an  extei'ior  point. 


434 


GEOMETRY, 


BOOK    IX. 


Proposition  XXXIII.     Theorem. 

914.  The  tangents  to  an  hyperbola  drawn  from  the 
centre  r)xeet  the  curve  at  an  infinite  distance  from 
the  centre. 


\ 


Let  OR  be  the  tangent  from  0. 

To  prove  that  OR  meets  the  curve  at  an  infinite  distance. 

Proof.   Let  G  be  the  intersection  of  arcs  described  from  0 
ftnd  -F'  as  centres  with  Oi^and  2  a  as  radii. 

The  point  of  contact  is  the  intersection  of  F^G  and  OR.  §  912 
Join  FG,  cutting  OR  at  Q. 

Now  OF'  -  0F\     ff 

also  QG=QF.  §910 

Therefore  F'G  is  D  to  OR.  §  189 

Therefore  the  point  of  contact  is  at  an  infinite  distance. 

Q.  E.  D 


THE   HYPERBOLA.  435 

In  the  same  way  another  tangent  OS  may  be  drawn,  meet- 
ing the  other  branch  of  the  curve  at  an  infinite  distance. 

915.  The  lines  OB,  OS.  indefinitely  produced  in  both  direc- 
tions, are  called  the  asymptotes  of  the  hyperbola. 

916.  Cor.  1.  ITie  line  FG  is  tangent  to  the  auxiliary  circle 
at  Q. 

For  FG  is  ±\o  OB,  §  910 

Therefore  Q  lies  on  the  auxiliary  circle.  §  91 1 

Hence  FG  touches  the  auxiliary  circle  at  Q.  §  239 

917.  Cor.  2.  FQ  is  equal  to  the  semi-conjugate  axis  b. 

For  F^  =  OF'-0^,                          §339 

and  6«  =  c»-a*.                                 §890 

But  0F=  c,  and  OQ  =  a. 

Therefore  FQ  =  b, 

918.  Cor.  3.  If  the  tangent  to  the  curve  at  A  meets  the 
asymptote  OB  at  B,  then  AB  =  b. 

For  A  OAB  =  A  OQF  §  149 

Therefore  AB  =  FQ  =  b. 

919.  Cob.  4.  The  asymptotes  of  an  hyperbola  are  the  diago- 
nals of  the  rectangle  BSUV  constructed  with  0  for  its  centre, 
and  the  transverse  and  conjugate  axes  for  its  two  sides. 

920.  A  perpendicular  to  a  tangent  erected  at  the  point  oi 
contact  is  called  a  normal. 

The  terms  subtangent  and  subnormal  are  used  in  the  hyper- 
bola in  the  same  sense  as  in  the  ellipse.  §  872 


436 


GEOMETRY.  —  BOOK   IX. 


Proposition  XXXIV.     Theorem. 

921.  The  section  of  a  right  circular  cone  made  hy  a 
plane  that  cuts  both  nappes  of  the  cone  is  an  hyper- 
bola. 


Q 


N'l 


T). 

/t\ 

x 

--  JK                    ...   '\ 

P             0 

Q 


m 


Let  a  plane  cut  the  lower  nappe  of  the  cone  in  the 
curve  FAQ,  and  the  upper  nappe  in  the  curve  P'A'Q'. 

To  prove  that  FAQ  and  F'A'Q'  are  the  two  branches  of  an 
hyperbola.  0 

Proof.  The  plane  passed  through  the  axis  of  the  cone  per- 
pendicular to  the  secant  plane  cuts  the  surface  of  the  cone  in 
the  elements  BS,  CS  (prolonged  through  S),  and  the  secant 
plane  in  the  line  UN', 


THE   HYPERBOLA.  437 

Describe  the  ©  0,  0',  tangent  to  BS,  OS,  JSTJV'.  Let  the 
points  of  contact  be  D,  IT,  F,  and  D',  H',  F\  respectively. 

Turn  B8C  and  the  (D  0  and  0'  about  the  axis  of  the  cone. 
B8  and  C8  will  generate  the  surfaces  of  the  two  nappes  of  a 
right  circular  cone ;  and  the  ©  0,  0'  will  generate  spheres 
which  touch  the  cone  in  the  ©  DKH,  D'K'II\  and  the 
secant  plane  in  the  points  F,  FK 

Let  P  be  any  point  on  the  curve.  Draw  PF  and  PF^ ; 
and  draw  P8  which  touches  the  ©  DKH,  B^K^W  at  the 
points  K.  K'. 

Now  PjFand  P^^are  tangents  to  the  sphere  0  from  the 
point  P. 

Therefore  PF=  PK. 

Also  PF'  =  PK'. 

Hence  PF'  -  PF^  PK'  -  PK 

■=  KK',  a  constant  qunntity. 

Therefore  the  curve  is  an  hyperbola  with  the  poinrs  F  and 
F'  for  foci.  * 


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